Mathematics for the International Student:Mathematics HL options: Statistics and Probability for IB, HL Topic 7, FM Topic 3 1921972319, 9781921972317

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HAESE MATHEMATICS Specialists in mathematics publishing

Mathematics

for the international student

Mathematics HL (Option): Statistics and Probability

HL Topic 7 FM Topic 3 Catherine Quinn Peter Blythe Robert Haese Michael Haese

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for use with IB Diploma Programme black

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MATHEMATICS FOR THE INTERNATIONAL STUDENT Mathematics HL (Option): Statistics and Probability Catherine Quinn Peter Blythe Robert Haese Michael Haese

B.Sc.(Hons), Grad.Dip.Ed., Ph.D. B.Sc. B.Sc. B.Sc.(Hons.), Ph.D.

Haese Mathematics 152 Richmond Road, Marleston, SA 5033, AUSTRALIA Telephone: +61 8 8210 4666, Fax: +61 8 8354 1238 Email: [email protected] www.haesemathematics.com.au Web: National Library of Australia Card Number & ISBN 978-1-921972-31-7 © Haese & Harris Publications 2013 Published by Haese Mathematics. 152 Richmond Road, Marleston, SA 5033, AUSTRALIA First Edition

2013

Artwork by Brian Houston. Cover design by Piotr Poturaj. Typeset in Australia by Deanne Gallasch. Typeset in Times Roman 10 \Qw_ . Printed in Malaysia through Bookpac Production Services, Singapore.

The textbook and its accompanying CD have been developed independently of the International Baccalaureate Organization (IBO). The textbook and CD are in no way connected with, or endorsed by, the IBO. This book is copyright. Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, research, criticism or review), no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher. Enquiries to be made to Haese Mathematics. Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers it has given a remuneration notice to Copyright Agency Limited (CAL). For information, contact the Copyright Agency Limited. Acknowledgements: While every attempt has been made to trace and acknowledge copyright, the authors and publishers apologise for any accidental infringement where copyright has proved untraceable. They would be pleased to come to a suitable agreement with the rightful owner.

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Disclaimer: All the internet addresses (URLs) given in this book were valid at the time of printing. While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher.

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FOREWORD Mathematics HL (Option): Statistics and Probability has been written as a companion book to the Mathematics HL (Core) textbook. Together, they aim to provide students and teachers with appropriate coverage of the two-year Mathematics HL Course, to be first examined in 2014. This book covers all sub-topics set out in Mathematics HL Option Topic 7 and Further Mathematics HL Topic 3, Statistics and Probability. The aim of this topic is to introduce students to the basic concepts and techniques of statistics and probability and their applications. Detailed explanations and key facts are highlighted throughout the text. Each sub-topic contains numerous Worked Examples, highlighting each step necessary to reach the answer for that example. Theory of Knowledge is a core requirement in the International Baccalaureate Diploma Programme, whereby students are encouraged to think critically and challenge the assumptions of knowledge. Discussion topics for Theory of Knowledge have been included on pages 157 to 159. These aim to help students discover and express their views on knowledge issues. The accompanying student CD includes a PDF of the full text and access to specially designed graphing software. Graphics calculator instructions for Casio fx-9860G Plus, Casio fx-CG20, TI-84 Plus and TI-nspire are available from icons located throughout the book. Fully worked solutions are provided at the back of the text, however students are encouraged to attempt each question before referring to the solution. It is not our intention to define the course. Teachers are encouraged to use other resources. We have developed this book independently of the International Baccalaureate Organization (IBO) in consultation with experienced teachers of IB Mathematics. The Text is not endorsed by the IBO. In this changing world of mathematics education, we believe that the contextual approach shown in this book, with associated use of technology, will enhance the students understanding, knowledge and appreciation of mathematics and its universal applications. We welcome your feedback. Email:

[email protected]

Web:

www.haesemathematics.com.au

CTQ

PJB

RCH PMH

ACKNOWLEDGEMENTS

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The authors and publishers would like to thank all those teachers who offered advice and encouragement on this book.

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USING THE INTERACTIVE STUDENT CD The interactive CD is ideal for independent study. Students can revisit concepts taught in class and undertake their own revision and practice. The CD also has the text of the book, allowing students to leave the textbook at school and keep the CD at home. By clicking on the relevant icon, a range of interactive features can be accessed: w Graphics calculator instructions for the Casio fx-9860G Plus, Casio fxCG20, TI-84 Plus and the TI-nspire w Interactive links to graphing software

INTERACTIVE LINK

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GRAPHICS CALCUL ATOR INSTRUCTIONS

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TABLE OF CONTENTS

5

TABLE OF CONTENTS

SYMBOLS AND NOTATION USED IN THIS BOOK Expectation algebra Discrete random variables Continuous random variables Probability generating functions Distributions of the sample mean and the Central Limit Theorem Point estimation (unbiased estimators and estimates) Confidence intervals for means Significance and hypothesis testing Bivariate Statistics

9 26 42 52 66 82 90 100 124

Review set A Review set B Review set C Review set D

146 148 151 153

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INDEX

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160

75

WORKED SOLUTIONS

25

158

0

THEORY OF KNOWLEDGE (Population Parameters)

5

157

95

THEORY OF KNOWLEDGE (The Central Limit Theorem)

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A B C D E F G H I

6

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6

SYMBOLS AND NOTATION USED IN THIS BOOK ¼ > > < 6 f......g 2 2 = N Z Q R Z+ µ ½ ) ) Á f : A!B f : x 7! y f (x) f ±g jxj [a, b] ] a, b [ un fun g Sn S1 n P ui i=1 n Q

is approximately equal to is greater than is greater than or equal to is less than is less than or equal to the set of all elements ...... is an element of is not an element of the set of all natural numbers f0, 1, 2, 3, ....g the set of integers f0, §1, §2, §3, ....g the set of rational numbers the set of real numbers the set of positive integers f1, 2, 3, ....g is a subset of is a proper subset of implies that does not imply that f is a function under which each element of set A has an image in set B f is a function under which x is mapped to y the image of x under the function f or f (g(x)) the composite function of f and g the modulus or absolute value of x the the the the the the

closed interval a 6 x 6 b open interval a < x < b nth term of a sequence or series with first term u1 sequence with nth term un , if first term is u1 sum of the first n terms of a sequence sum to infinity of a convergent series

u1 + u2 + u3 + :::: + un u1 £ u2 £ u3 £ :::: £ un

ui

i=1

lim f (x)

the limit of f(x) as x tends to a

lim f(x)

the limit of f(x) as x tends to a from the positive side of a

x!a

x!a+

lim f(x) the limit of f(x) as x tends to a from the negative side of a

x!a¡

maxfa, bg 1 P cn xn

the maximum value of a or b the power series whose terms have form cn xn

n=0

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the derivative of y with respect to x

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dy dx

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7 f 0 (x)

the derivative of f (x) with respect to x

2

d y dx2 f 00 (x) dn y dxn (n) f (x) R y dx Rb y dx a

the second derivative of y with respect to x the second derivative of f (x) with respect to x the nth derivative of y with respect to x the nth derivative of f(x) with respect to x the indefinite integral of y with respect to x

the definite integral of y with respect to x between the limits x = a and x = b ex exponential function of x ln x the natural logarithm of x sin, cos, tan the circular functions csc, sec, cot the reciprocal circular functions arcsin, arccos, arctan the inverse circular functions ¡n¢ n! r!(n ¡ r)!

r

P(A)

probability of event A

P(A0 )

probability of the event “not A” probability of the event A given B observations probability distribution function Px = P(X = x) of the discrete random variable X probability density function of the continuous random variable X cumulative distribution function of the continuous random variable X the expected value of the random variable X the variance of the random variable X population mean n P

P(A j B) x1 , x2 , .... P(x) f (x) F (x) E(X) Var(X) ¹ ¾2

population variance, the value ¾ 2 =

¾ x

population standard deviation sample mean

n P

(xi ¡ x)2

sample variance, the value

sn

standard deviation of the sample of size n

=

, for a population of size n

n

sn2 sn2¡1

sn2

(xi ¡ ¹)2

i=1

i=1

n

, from a sample of size n

unbiased estimate of the population variance, the value n P 2 sn2¡1 =

(xi ¡ x) n sn2 = i=1 , from a sample of size n n¡1 n¡1

the estimator of ¹, that is the function X =

X

n 1 P Xi , where Xi , i = 1, ...., n n i=1

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are identically distributed independent random variables each with mean ¹

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8 n P

Sn2

2

the biased estimator of ¾ , that is the function

Sn2

=

(Xi ¡ X)2

i=1

n

where Xi , i = 1, ...., n are identically distributed independent random variables each with variance ¾ 2 Sn2¡1

the unbiased estimator of ¾ 2 , that is the function Sn2¡1 =

DU(n) B(1, p) B(n, p)

n S2 n¡1 n

discrete uniform distribution with parameter n Bernoulli distribution with parameter p binomial distribution with parameters n and p

Geo(p) NB(r, p) Po(m) X » DU(n)

geometric distribution with parameter p negative binomial distribution with parameters r and p Poisson distribution with mean m the random variable X has a discrete uniform distribution with parameter n

X » B(1, p) X » B(n, p) X » Geo(p)

the random variable X has a Bernoulli distribution with parameter p the random variable X has a binomial distribution with parameters n and p the random variable X has a geometric distribution with parameter p

X » NB(r, p) the random variable X has a negative binomial distribution with parameters r and p X » Po(m) the random variable X has a Poisson distribution with mean m U(a, b) continuous uniform distribution with parameters a and b Exp(¸)

exponential distribution with mean

1 ¸

N(¹, ¾ 2 ) º t(º) X » U(a, b)

normal distribution with mean ¹ and variance ¾ 2 number of degrees of freedom Student’s t-distribution with º degrees of freedom the random variable X has a continuous uniform distribution with parameters a and b

X » Exp(¸)

the random variable X has an exponential distribution with mean

1 ¸

X » N(¹, ¾ 2 ) the random variable X has a normal distribution with mean ¹ and variance ¾ 2 T » t(º) the random variable T has the Student’s t-distribution with º degrees of freedom

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pb H0 H1 ® ¯ 1¡¯ Cov(X, Y ) ½ R r

75

p

25

the probability generating function E(tX ) for a discrete random variable X which takes values in N depending on the context, a parameter of a distribution, a population proportion, or a p-value in a hypothesis test a sample proportion null hypothesis alternative hypothesis significance level or probability of a Type I error probability of a Type II error power of a hypothesis test covariance of random variables X and Y product moment correlation coefficient between two random variables the sample product moment correlation coefficient; an estimator of ½ the observed value of R for a given sample of bivariate data; an estimate of ½

G(t)

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9

STATISTICS AND PROBABILITY

A

EXPECTATION ALGEBRA

A random variable can take any one of a set of values from a given domain, according to given probabilities. The domain may be discrete or continuous.

DISCRETE RANDOM VARIABLES If X is a discrete random variable, then: 1 X has possible values x1 , x2 , x3 , .... . To determine the value of X we usually count. P pi = 1. 2 X takes value xi with probability pi , where 0 6 pi 6 1, i = 1, 2, 3, ...., and 3 X has a probability distribution function (or probability mass function) P (x), where P (xi ) = P(X = xi ) = pi , i = 1, 2, 3, .... . P 4 X has a cumulative distribution function (CDF) F (x), where P P(X = xi ). F (x) = P(X 6 x) =

xi 6 x

xi 6 x

reads “the sum for all values of xi less than or equal to x”.

F (x) is the probability that X takes a value less than or equal to x. Examples of discrete probability distributions and random variables covered in the Core course include Bernoulli, Discrete Uniform, Binomial, and Poisson.

CONTINUOUS RANDOM VARIABLES If X is a continuous random variable, then: 1 The possible values of X may be all x 2 R , or all real x in some domain [a, b]. To determine the value of X we usually measure. 2 X has a continuous probability density function (PDF) f (x), where: ² f (x) > 0 for all x in the domain of f . Z 1 f(x) dx = 1 if the domain of f is R , or ²

y y = f(x) Area = 1

¡1

Z

b

f(x) dx = 1 if the domain of f is [a, b].

a

x

3 Suppose f and the PDF for X, have domain [a, b]. X has a cumulative distribution function Z x f (t) dt for x 2 [a, b]. (CDF) F (x), where F (x) = P(X 6 x) = a

² F (x) is the probability that X takes a value less than or equal to x. Z b f(t) dt = 1 ² F (b) =

y y = f(t) Area = F(x)

a

x b

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t

10

STATISTICS AND PROBABILITY

² The probability that X takes a value in the interval [c, d] µ [a, b] is given by Z d f (t) dt P(c 6 X 6 d) =

y Area = P(c 6 X 6 d)

c

= F (d) ¡ F (c) a

c

d b

t

4 Since X has infinitely many possible values, the probability that X takes a single value X = x is 0. However, since X is a continuous random variable, for x 2 Z and x ¡ 0:5 6 X < x + 0:5, the value of X will be rounded to the integer x. Thus, for x 2 Z , we define P(X = x) = P(x ¡ 0:5 6 X < x + 0:5) Z x+0:5 = f(t) dt x¡0:5

= F (x + 0:5) ¡ F (x ¡ 0:5)

You should recognise the Normal distribution from the Core course.

Example 1 Given a random variable X » N(7:2, 28), find P(X = 10). P(X = 10) = P(9:5 6 X < 10:5) ¼ 0:0655 P(c 6 X < d)

For a continuous random variable X,

= P(c 6 X 6 d) = P(c < X 6 d) = P(c < X < d) since the corresponding integrals all define the same area under the curve y = f(t) between t = c and t = d. Examples of continuous probability distributions and random variables covered in the Core course include Continuous Uniform, Exponential, and the Normal distribution.

EXPECTATION The mean or expected value or expectation E(X) of a random variable X is defined as follows: ² If X is a discrete random variable with set of possible values x1 , x2 , .... and probability mass P function P(X = xi ) = pi , i = 1, 2, .... , E(X) = ¹ = xi P(X = xi ) i P = xi pi , i = 1, 2, .... i

² If X is a continuous random variable with probability density function f(x) with domain [a, b], Z b x f (x) dx. E(X) = ¹ =

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STATISTICS AND PROBABILITY

Example 2 Find the expectation of the following random variables: x

1

2

P(X = x)

2 3

1 3

a a discrete random variable X with probability distribution: b the continuous uniform random variable X » U(1, 4). a E(X) = 1 £ = 43

2 3

+2£

Z

1 3

b E(X) =

· = = =

4

1

1 3

x£

dx

¸4

x2 6 1

15 6 5 2

Theorem 1 1 E(d) = d for any constant d. 2 E(cX + d) = cE(X) + d for X a random variable and constants c, d 2 R . 1 A constant d takes value d with probability 1. ) by definition, E(d) = d £ 1 = d.

Proof:

2 Discrete Case: X takes value xi with probability pi . Let U = cX + d be a new random variable. ) U takes value cxi + d with probability pi , i = 1, 2, .... . ) by the definition of E(U ), P E(U ) = E(cX + d) = (cxi + d) pi P = (cxi pi + dpi ) P P xi pi + d pi =c P = cE(X) + d fsince pi = 1g Continuous Case: If X has PDF f (x) with domain [a, b], then Z b (cx + d) f(x) dx E(cX + d) = a

Z

b

=c

Z

b

x f(x) dx + d

a

f(x) dx a

Z

b

= cE(X) + d fsince

f (x) dx = 1g

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STATISTICS AND PROBABILITY

Example 3 Consider a 6-sided unbiased die with face labels: ¡2, ¡2, 0, 1, 1, 2. Let X be the discrete random variable with possible values the outcome of a roll of the die. a Find E(X) and E(X 2 ).

b Show that E(X 2 ) 6= (E(X))2 .

a The discrete random variable X has probability distribution: 1 3

) E(X) = ¡2 £

+0£ 2

1 6

1 3

+1£

x

¡2

0

1

2

P(X = x)

1 3

1 6

1 3

1 6

+2£

1 6

=0

Consider the variable X which has possible values 0, 1, and 4. X 2 has probability distribution:

X2

0

1

4

Probability

1 6

1 3

1 2

P(X 2 = 4) = P(X = ¡2 or 2) = =

) E(X 2 ) = 0 £ b E(X 2 ) =

7 3

1 6

+1£

1 3 2

+4£

1 2

=

1 3 1 2

+

1 6

7 3

6= 0 = (E(X))

In Example 3 we see that E(X 2 ) = 02 £ = 02 £ 4 P

=

i=1

1 6 1 6

+ 12 £ + 12 £

1 3 1 3

+4£

¡1 3 2

+ (¡2) £

¢

1 6 1 3 +

+

22 £

1 6

xi2 pi

where X has values x1 = ¡2, x2 = 0, x3 = 1, x4 = 2 with probabilities p1 = 13 , p2 = 16 , p3 = 13 , p4 = 16 respectively. P ) E(g(x)) = g(xi ) pi , where g(X) = X 2 . This result generalises for any function g. Theorem 2 Suppose X is a random variable and g is any function. The random variable g(X) has mean given by: P Discrete case: E(g(X)) = g(xi ) pi Z Continuous case: E(g(X)) = g(x) f (x) dx The proof for the continuous case is analogous.

Proof for the discrete case: g(X) has values g(x1 ), g(x2 ), .... which are not necessarily distinct.

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g(X) takes value g(xi ) with probability pi , i = 1, 2, 3, ::::, or if g(x) = g(x1 ) = :::: = g(xk ) then g(X) takes value g(x) with probability p1 + p2 + :::: + pk . P Hence, from the definition of expected value, E(g(x)) = g(xi ) pi .

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STATISTICS AND PROBABILITY

For X a random variable, E(cg(X) § dh(X)) = c E(g(X)) § d E(h(X))

Corollary:

where c, d are constants and g(x) and h(x) are functions. Proof for the discrete case: By Theorem 1 and Theorem 2,

E(cg(X) § dh(X)) P = (cg(xi ) § dh(xi )) pi P P =c g(xi ) pi § d h(xi ) pi = c E(g(X)) § d E(h(X))

VARIANCE The variance ¾ 2 , also denoted Var(X), of a random variable X is ¾ 2 = Var(X) = E((X ¡ ¹)2 ), where ¹ = E(X). Var(X) is the mean of the squared differences of values of X from the mean value of X, E(X). Var(X) is therefore a measure of the spread of the distribution of X. Theorem 3 For X a random variable, Var(X) = E(X 2 ) ¡ fE(X)g2 . Var(X) = E((X ¡ ¹)2 ) = E(X 2 ¡ 2X¹ + ¹2 ) = E(X 2 ) ¡ E(2¹X) + E(¹2 ) = E(X 2 ) ¡ 2¹E(X) + ¹2 = E(X 2 ) ¡ 2¹2 + ¹2 = E(X 2 ) ¡ ¹2 = E(X 2 ) ¡ fE(X)g2

Proof:

Using Theorem 3, we find that: Var(X) =

Discrete case:

¡P Z

Continuous case: Var(X) =

fby the Corollary to Theorem 2g fby Theorem 1g

¢ xi2 pi ¡ ¹2

x2 f(x) dx ¡ ¹2

Theorem 4 For X a random variable and constants c, d 2 R : 1 Var(d) = 0 2 Var(cX + d) = c2 Var(X) Proof:

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fby Theorem 3g fby Theorem 1 since d2 , d are constantsg

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1 For any constant d, Var(d) = E(d2 ) ¡ fE(d)g2 = d2 ¡ d2 =0

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13

14

STATISTICS AND PROBABILITY

2 Var(cX + d) = E((cX + d)2 ) ¡ fE(cX + d)g2 = E(c2 X 2 + 2cdX + d2 ) ¡ fcE(X) + dg2 = c2 E(X 2 ) + 2cd E(X) + E(d2 ) ¡ fc2 (E(X))2 + 2cd E(X) + d2 g = c2 E(X 2 ) + d2 ¡ c2 (E(X))2 ¡ d2 = c2 fE(X 2 ) ¡ fE(X)g2 g = c2 Var(X)

fby Theorem 3g

LINEAR TRANSFORMATION OF A SINGLE RANDOM VARIABLE Let X be a random variable. Let U be a new random variable obtained from X by the linear transformation U = cX + d, where c, d are constants. By Theorems 1 and 4 we have: E(U ) = E(cX + d) = cE(X) + d and Var(U ) = Var(cX + d) = c2 Var(X) Theorem 5 Let X be a random variable with mean ¹ and standard deviation ¾. Then X ¤ = X¤ =

X ¡¹ is a random variable with E(X ¤ ) = 0 and Var(X ¤ ) = 1. ¾

X ¡¹ is called the standardised variable corresponding to X. ¾

³

´

1 ¹ X¡ ¾ ¾ 1 ¹ = E(X) ¡ ¾ ¾ 1 ¹ = £¹¡ ¾ ¾

E(X ¤ ) = E

Proof:

Var(X ¤ ) = Var

and

³

1 ¹ X¡ ¾ ¾

´

1 Var(X) ¾2 1 = 2 £ ¾2 ¾

=

=0

=1

For example, if X » N(¹, ¾ 2 ) is a continuous random variable with a normal distribution with mean ¹ and variance ¾ 2 , then Z =

X ¡¹ has E(Z) = 0 and Var(Z) = 1. ¾

Z » N(0, 1), where N(0, 1) is the standard normal distribution, as studied in the Core course.

DEFINING NEW RANDOM VARIABLES FROM OLD In the following example we use two random variables X and Y to define new random variables 3X ¡5, X 2 , X + Y , and XY .

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We observe how the mean and variance of the new random variables are related to the mean and variance of X and Y , as shown in Theorems 1 - 4.

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STATISTICS AND PROBABILITY

15

Example 4 X and Y are independent random variables.

2

Let X be the random variable with value the outcome of spinning the equilateral triangle spinner:

1 1

3

2

-2

2

2

Let Y be the random variable with value the outcome of spinning the regular pentagon spinner:

a Determine the probability distributions for each of: i X and X 2 ii Y and Y 2 iv X + Y and (X + Y )2

iii 3X ¡ 5 and (3X ¡ 5)2

v XY and (XY )2

b Find the mean and variance of: i X ii Y iv X + Y v XY c Verify that: i ii iii iv

a

iii 3X ¡ 5

fE(X)g2 6= E(X 2 ) E(3X ¡ 5) = 3E(X) ¡ 5 and Var(3X ¡ 5) = 32 £ Var(X) E(X + Y ) = E(X) + E(Y ) and Var(X + Y ) = Var(X) + Var(Y ) E(XY ) = E(X) E(Y )

i X has probability distribution:

X

1

2

Probability

2 3

1 3

) X 2 has probability distribution:

X has values 1, 2 ) X 2 has values 12 = 1, 22 = 4 X2

1

4

Probability

2 3

1 3

ii Y has probability distribution: Y

¡2

2

3

Probability

1 5

3 5

1 5

P(Y 2 = 9) = P(Y = 3) =

Y has values ¡2, 2, 3 ) Y 2 has values (¡2)2 = 4, 22 = 4, 32 = 9

1 5

P(Y 2 = 4) = P(Y = ¡2 or Y = 2)

and

= =

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) Y 2 has probability distribution:

1 5 4 5

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16

STATISTICS AND PROBABILITY

iii

X 3X ¡ 5

1 ¡2

2 1

(3X ¡ 5)2

4

1

Probability

2 3

1 3

iv The possible outcomes of X + Y are: X +Y 1 X 2 P((X + Y ) = ¡1)

Y 2 3 4

¡2 ¡1 0

3 4 5

P((X + Y ) = 4)

and

= P(X = 1 and Y = ¡2)

= P(X = 2 and Y = 2) + P(X = 1 and Y = 3)

= P(X = 1) £ P(Y = ¡2)

=

= = )

2 1 3 £ 5 2 15

=

1 3 3 £ 5 5 15

+

2 3

£

1 5

fsince X and Y are independentg

and similarly for the remaining values. X +Y

¡1

0

3

4

5

Probability

2 15

1 15

6 15

5 15

1 15

X + Y has probability distribution:

) (X + Y )2 has probability distribution:

(X + Y )2

1

0

9

16

25

Probability

2 15

1 15

6 15

5 15

1 15

v The possible values of XY are: ¡2 ¡2 ¡4

XY 1 X 2

Y 2 2 4

3 3 6

P(XY = 3) = P(X = 1 and Y = 3) = P(X = 1) £ P(Y = 3) = =

2 1 3 £ 5 2 and 15

fsince X and Y are independentg

similarly for the remaining values. XY

¡4

¡2

2

3

4

6

Probability

1 15

2 15

6 15

2 15

3 15

1 15

) XY has probability distribution:

) (XY )2 has values (¡4)2 = 16, (¡2)2 = 4, 22 = 4, 32 = 9, 42 = 16, 62 = 36. P((XY )2 = 16) = P(XY = ¡4 or XY = 4) = P(XY = ¡4) + P(XY = 4) +

3 15

and similarly for the remaining values.

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) (XY )2 has probability distribution:

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=

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=

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17

STATISTICS AND PROBABILITY

i E(X) = 1 £

b

2 3

1 3

+2£

=

4 3

ii E(Y ) = ¡2 £

Var(X) = E(X 2 ) ¡ fE(X)g2 =2¡ =

2 9

+2£

3 5

+3£

Var(Y ) = E(Y ) ¡ fE(Y )g2

+ 4 £ 13 g ¡ f 43 g2

2 3 16 9

= f1 £

1 5 2

4 5 49 25

= f4 £ =5¡ 76 25

=

1 5

+ 9 £ 15 g ¡ f 75 g2

iii E(3X ¡ 5) = ¡2 £ 23 + 1 £ 13 = ¡1 Var(3X ¡ 5) = E((3X ¡ 5)2 ) ¡ fE(3X ¡ 5)g2 + 1 £ 13 g ¡ f¡1g2

2 3

= f4 £ =2 iv E(X + Y ) = ¡1 £

2 15

+0£

1 15

6 15

+3£

5 15

+4£

= 41 15 Var(X + Y ) = E((X + Y )2 ) ¡ fE(X + Y )g2 = f1 £ 734 225 1 £ 15

2 15

+0£

1 15

+9£

6 15

+ 16 £

+5£

5 15

1 15

+ 25 £

1 15 g

2 ¡ f 41 15 g

= v E(XY ) = ¡4

¡2£

2 15

+2£

6 15

2 15

+3£

= 28 15 Var(XY ) = E((XY )2 ) ¡ fE(XY )g2 = f4 £ =

1466 225

8 15

+9£

2 15

+ 16 £

4 15

3 15

+4£

+ 36 £

1 15 g

ii E(3X ¡ 5) = ¡1 and 3E(X) ¡ 5 = 3 £ Hence E(3X ¡ 5) = 3E(X) ¡ 5.

4 3

4 3

+

2 9

= 2.

7 5

=

Var(X + Y ) = 734 225 and Var(X) + Var(Y ) = Hence Var(X + Y ) = Var(X) + Var(Y ). iv E(XY ) = 28 15 and E(X)E(Y ) = Hence E(XY ) = E(X)E(Y ).

4 3

£

7 5

=

16 9

6= 2.

¡ 5 = ¡1.

Var(3X ¡ 5) = 2 and 32 £ Var(X) = 9 £ Hence Var(3X ¡ 5) = 32 Var(X). iii E(X + Y ) = 41 15 and E(X) + E(Y ) = Hence E(X + Y ) = E(X) + E(Y ).

1 15

2 ¡ f 28 15 g

i E(X 2 ) = 1 £ 23 + 4 £ 13 = 2 and fE(X)g2 = f 43 g2 = Hence E(X 2 ) 6= fE(X)g2

c

+6£

2 9

41 15 .

+

76 25

=

734 225 .

28 15 .

We now summarise and prove the results observed in the previous example. Theorem 6 For X, Y two random variables (either both discrete or both continuous): 1 E(X + Y ) = E(X) + E(Y ) 2 If X and Y are independent, then

Var(X + Y ) = Var(X) + Var(Y )

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and E(XY ) = E(X)E(Y )

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7 5

18

STATISTICS AND PROBABILITY

Proof for the discrete case: 1

PP

E(X + Y ) PP = (x + y) P(X = x and Y = y) x

y

PP

=

x

xP(X = x and Y = y) +

y

x

PP x

y

is used to cover all possible outcomes for X added to all possible outcomes for Y.

yP(X = x and Y = y)

y

P P P P x P(X = x and Y = y) + y P(X = x and Y = y)

=

x

y

y

fx is constant whilst summing over y.g =

P

xP(X = x) +

P

x

x

fBy first changing the order of the sum, we notice y is constant whilst summing over x.g yP(Y = y)

fIn a joint probability distribution, P P(X = x and Y = y) P(X = x) =

y

all y

P

and P(Y = y) =

P(X = x and Y = y)

all x

This is equivalent to summing a whole row or whole column in an array of a joint probability distribution.g = E(X) + E(Y ) 2 For X and Y independent, P(X = x and Y = y) = P(X = x)P(Y = y) PP E(XY ) = xyP(X = x and Y = y) = =

=

x

y

x

P

y

P xP(X = x) yP(Y = y)

x

y

PP

P

xyP(X = x)P(Y = y)

fx and P(X = x) are constants while summing over yg xP(X = x)E(Y )

x

= E(Y )

P

xP(X = x) fE(Y ) is a constant while summing over xg

x

= E(Y )E(X) = E(X)E(Y ) Var(X + Y ) = E((X + Y )2 ) ¡ fE(X + Y )g2 = E(X 2 + 2XY + Y 2 ) ¡ fE(X) + E(Y )g2 2

fBy part 1 of Theorem 6g

2

= E(X ) + 2E(XY ) + E(Y )

fCorollary to Theorem 2g

2

2

¡ ffE(X)g + 2E(X)E(Y ) + fE(Y )g g

fusing E(XY ) = E(X)E(Y ) from aboveg

= E(X 2 ) ¡ fE(X)g2 + E(Y 2 ) ¡ fE(Y )g2 | {z } | {z }

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Var(Y )

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STATISTICS AND PROBABILITY

19

In the following example we show that for X and Y dependent random variables either E(XY ) 6= E(X)E(Y ) or E(XY ) = E(X)E(Y ), depending on the example. Example 5 For each of the following distributions for X, let Y = X 2 be a random variable dependent on X. Compare E(XY ) with E(X)E(Y ). x

¡1

0

1

P(X = x)

1 3

1 3

1 3

a

a E(X) = ¡1 £

1 3

+0£

1 3

x

1

2

3

P(X = x)

1 3

1 3

1 3

b

+1£

1 3

=0

Let Y = X 2 , so Y is dependent on X. P(X 2 = 1) = P(X = 1 or X = ¡1) = =

1 3 2 3

+

1 3

) Y has probability distribution:

) E(Y ) = E(X 2 ) = 0 £

1 3

+1£

2 3

Y = X2

0

1

Probability

1 3

2 3

XY = X 3

¡1

0

1

Probability

1 3

1 3

1 3

2 3

=

Consider XY = X 3 which has values ¡1, 0, 1 XY has probability distribution:

) E(XY ) = E(X 3 ) = ¡1 £

1 3

+0£

1 3

+1£

Hence E(XY ) = 0 and E(X)E(Y ) = 0 £

2 3

1 3

=0

=0

Thus E(XY ) = E(X)E(Y ) even though X and Y are dependent. b E(X) = 1 £

1 3

+2£

1 3

+3£

1 3

=2

Let Y = X 2 which has values 1, 4, 9 Y

1

4

9

Probability

1 3

1 3

1 3

) E(Y ) = 1 £

1 3

+4£

1 3

+9£

1 3

=

14 3

Consider XY = X 3 which has values 1, 8, 27 X3

1

8

27

Probability

1 3

1 3

1 3

) E(X 3 ) = 1 £

1 3

Hence E(XY ) = E(X 3 ) = 12 but E(X)E(Y ) = 2 £

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20

STATISTICS AND PROBABILITY

We now summarise the main results for this section which we obtain from the theorems above and by Mathematical Induction: Theorem 7 For X1 , X2 , ...., Xn random variables (either all discrete or all continuous) and a1 , a2 , ...., an constants: 1 E(a1 X1 § a2 X2 § :::: § an Xn ) = a1 E(X1 ) § a2 E(X2 ) § :::: § an E(Xn ) 2 If X1 , X2 , ...., Xn are independent random variables, then E(X1 X2 ::::Xn ) = E(X1 )E(X2 ) :::: E(Xn ) and Var(a1 X1 § a2 X2 § :::: § an Xn ) = a12 Var(X1 ) + a22 Var(X2 ) + :::: + an2 Var(Xn ) Proof: 1 By Theorem 1, E(§ai Xi ) = §ai E(Xi ) ) E(a1 X1 § a2 X2 § :::: § an¡1 Xn¡1 § an Xn ) = E(a1 X1 § a2 X2 § :::: § an¡1 Xn¡1 ) § an E(Xn ) = E(a1 X1 § a2 X2 § :::: § an¡2 Xn¡2 ) § an¡1 E(Xn¡1 ) § an E(Xn ) frepeated use of Theorem 6g .. . = a1 E(X1 ) § a2 E(X2 ) § :::: § an E(Xn ) 2 E(X1 X2 ::::Xn ) = E(X1 X2 ::::Xn¡1 )E(Xn ) = E(X1 X2 ::::Xn¡2 )E(Xn¡1 )E(Xn ) fletting Y = X1 X2 ::::Xn¡1 , X = Xn g .. . = E(X1 )E(X2 ):::: E(Xn )

frepeated use of Theorem 6g

ai2 Var(Xi )

By Theorem 4, Var(§ai Xi ) = By Theorem 6, letting X = a1 X1 § :::: § an¡1 Xn¡1 and Y = §an Xn , Var(a1 X1 § a2 X2 § :::: § an¡1 Xn¡1 § an Xn ) = Var(a1 X1 § a2 X2 § :::: § an¡1 Xn¡1 ) + an2 Var(Xn ) = Var(a1 X1 § :::: § an¡2 Xn¡2 ) + an2¡1 Var(Xn¡1 ) + an2 Var(Xn ) frepeated use of Theorem 6g .. . = a12 Var(X1 ) + :::: + an2 Var(Xn )

CONTINUOUS NORMALLY DISTRIBUTED RANDOM VARIABLES Theorem 8

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A linear combination of X1 , X2 , X3 has the form a1 X1 + a2 X2 + a3 X3 where a1 , a2 , a3 are constants.

Any linear combination of independent continuous normally distributed random variables is itself a continuous and normally distributed random variable.

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STATISTICS AND PROBABILITY

21

For example, if X1 , X2 , X3 are independent normal random variables then Y = 2X1 + 3X2 ¡ 4X3 is a normal random variable. By Theorem 7, E(Y ) = E(2X1 + 3X2 ¡ 4X3 ) = 2E(X1 ) + 3E(X2 ) ¡ 4E(X3 ) and Var(Y ) = Var(2X1 + 3X2 ¡ 4X3 ) = 4Var(X1 ) + 9Var(X2 ) + 16Var(X3 ) By Theorem 8, Y » N(E(Y ), Var(Y )). Example 6 The weights of male employees in a bank are normally distributed with mean ¹ = 71:5 kg and standard deviation ¾ = 7:3 kg. The bank has an elevator which will carry a maximum load of 444 kg. a Six male employees enter the elevator. Calculate the probability that their combined weight exceeds the maximum load. b If there is to be at most a 0:1% chance of the total weight exceeding 444 kg, recommend the maximum number of males who should use the lift together. a Let the weights of the employees be the independent random variables X1 , X2 , ...., X6 . We are concerned with the sum of their weights Y = X1 + X2 + X3 + X4 + X5 + X6 , where Xi » N(71:5, (7:3)2 ), i = 1, 2, ...., 6. Now E(Y ) = E(X1 ) + E(X2 ) + :::: + E(X6 ) = 6 £ 71:5 = 429 kg and

Var(Y ) = Var(X1 ) + Var(X2 ) + :::: + Var(X6 ) = 6 £ 7:32 = 319:74

) Y is normally distributed with mean 429 kg and variance 319:75 kg2 ) Y » N(429, 319:74) Now P(Y > 444) ¼ 0:201 So, there is a 20:1% chance that their combined weight will exceed 444 kg. b Six men is too many, as there is a 20:1% chance of overload. Instead we try 5 male employees with total weight Y = X1 + X2 + X3 + X4 + X5 . E(Y ) = 5 £ 71:5 Var(Y ) = 5 £ 7:32 = 357:5 kg ¼ 266:45 kg2

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Now Y » N(357:5, 266:45) ) P(Y > 444) ¼ 5:82 £ 10¡8 For n = 5, there is much less than a 0:1% chance of the total weight exceeding 444 kg. We recommend that a maximum of 5 men use the elevator together.

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22

STATISTICS AND PROBABILITY

Example 7 The random variable X has distribution with mean 11 and standard deviation 2. Define three independent random variables X1 = 2X, X2 = 4 ¡ 3X, and X3 = 4X + 1. Find the mean and standard deviation of the random variable (X1 + X2 + X3 ). Mean = E(X1 + X2 + X3 ) = E(X1 ) + E(X2 ) + E(X3 ) = 2E(X) + (4 ¡ 3E(X)) + (4E(X) + 1) = 3E(X) + 5 = 3(11) + 5 = 38

Variance = Var(X1 + X2 + X3 ) = Var(X1 ) + Var(X2 ) + Var(X3 ) = 4Var(X) + 9Var(X) + 16Var(X) = 29Var(X) = 29 £ 22 = 116

) the mean is 38 and the standard deviation is

p 116 ¼ 10:8 .

Example 8 A cereal manufacturer produces packets of cereal Mean (g) Variance (g2 ) in two sizes, small and economy. Small 315 4 The amount in each packet is distributed normally Economy 950 25 and independently with mean and variance as shown in the table. a A packet of each size is selected at random. Find the probability that the economy packet contains less than three times the amount in the small packet. b One economy and three small packets are selected at random. Find the probability that the economy packet contains less than the total amount in the three small packets. Let S be the weight of a small packet and E be the weight of an economy packet. ) S » N(315, 4) and E » N(950, 25). a The probability that the economy packet contains less than three times the amount in a small packet, is P(E < 3S) or P(E ¡ 3S < 0). Now

E(E ¡ 3S) = E(E) ¡ 3E(S) = 950 ¡ 3 £ 315 =5

Var(E ¡ 3S) = Var(E) + 9Var(S) = 25 + 9 £ 4 = 61

and

) E ¡ 3S » N(5, 61) ) P(E ¡ 3S < 0) ¼ 0:261 b The probability that the economy packet contains less than the total amount in the three small packets is P(E < S1 + S2 + S3 ) or P(E ¡ (S1 + S2 + S3 ) < 0), where S1 , S2 , S3 each have distribution N(315, 4). Now

E(E ¡ (S1 + S2 + S3 )) = E(E) ¡ 3E(S) = 950 ¡ 3 £ 315 =5

Var(E ¡ (S1 + S2 + S3 )) = Var(E) + Var(S1 ) + Var(S2 ) + Var(S3 ) = 25 + 12 = 37

and

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) E ¡ (S1 + S2 + S3 ) » N(5, 37) ) P(E ¡ (S1 + S2 + S3 ) < 0) ¼ 0:206

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23

STATISTICS AND PROBABILITY

EXERCISE A 1 Given a random variable X » N(5, 22), find: a P(X = 24)

b P(X = 30)

c P(X = 11)

2 Find the expectation of the following random variables: x

¡1

1

3

P(X = x)

1 2

1 3

1 6

a a discrete random variable X with probability distribution:

b the continuous uniform random variable X » U(0, 5). 3 Consider a 6-sided unbiased die with face labels ¡3, ¡1, ¡1, 2, 2, 3. Let X be the discrete random variable with possible values the outcome of a roll of the die. a Find E(X) and E(X 2 ).

b Show that E(X 2 ) 6= (E(X))2 .

4 Let X be the number of heads obtained when an unbiased coin is tossed twice. Let Y be the result when an ordinary unbiased 6-sided die is rolled once. a Write down the probability distributions for each of: i X and X 2 ii Y and Y 2 iii X + Y and (X + Y )2 iv 4X ¡ 2Y and (4X ¡ 2Y )2

v XY and (XY )2

b Find the mean and variance of: i X ii Y iii X + Y iv 4X ¡ 2Y v XY c Verify that: i E(X + Y ) = E(X) + E(Y ) and Var(X + Y ) = Var(X) + Var(Y ) ii E(4X ¡ 2Y ) = 4E(X) ¡ 2E(Y ) and Var(4X ¡ 2Y ) = 16Var(X) + 4Var(Y ) iii E(XY ) = E(X)E(Y ) 5 X and Y are two discrete random variables with E(X) = 3, E(Y ) = 2, Var(X) = Var(Y ) = 54 . Calculate, where possible:

3 2,

and

a E(X + Y ) b E(XY ) c Var(2X ¡ 3Y + 6) d E(5XY ) Hint: You need to consider the two cases: X, Y are independent and X, Y are dependent.

6 For each of the following distributions for X, let Y = X 2 be a random variable. Hence compare E(XY ) with E(X)E(Y ). x

0

1

2

3

P(X = x)

1 4

1 4

1 4

1 4

a

b

x

¡1

0

1

2

P(X = x)

1 4

1 4

1 4

1 4

7 Consider two independent normally distributed random variables X and Y with means and standard deviations shown. a Find the mean and standard deviation of 3X ¡ 2Y . b Find P(3X ¡ 2Y > 3). State clearly any results you use.

X Y

¹

¾

3:8 5:7

0:323 1:02

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8 The marks in an IB Mathematics HL exam are distributed normally with mean ¹ and standard deviation ¾. The cut off score for a 7 is a mark of 80%, and 10% of students get a 7. The cut off score for a 6 is a mark of 65%, and 20% of students get a 6. Find the mean and standard deviation of the marks in this exam.

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24

STATISTICS AND PROBABILITY

9 The maximum load of a lift is 440 kg. The weights of adults are normally distributed with mean 81 kg and standard deviation 11 kg. The weights of children are normally distributed with mean 48 kg and standard deviation 4 kg. Find the probability that if the lift contains 4 adults and 3 children then the maximum load will be exceeded. What assumption have you made in your calculation? 10 A coffee machine dispenses a cappuccino made up of black coffee distributed normally with mean 120 mL and standard deviation 7 mL, and froth distributed normally with mean 28 mL and standard deviation 4:5 mL. Each cup is marked to a level of 135:5 mL, and if this is not attained then the customer will receive their cappuccino free of charge. Determine whether or not the proprietor needs to adjust the settings on her machine if she wishes to give away no more than 1% of cappuccinos free. 11 X and Y are independent normal random variables with X » N(¡10, 1) and Y » N(25, 25). a Find the mean and standard deviation of the random varible U = 3X + 2Y . b Find P(U < 0). 12 A drinks manufacturer produces bottles of drink in two sizes, small (S) and large (L). The distributions for the contents are independent, and normally distributed with S » N(280 mL, 4 mL2 ) and L » N(575 mL, 16 mL2 ). a A bottle of each size is selected at random. Find the probability that the large bottle contains less than two times the amount in the small bottle. b One large and two small bottles are selected at random. Find the probability that the large bottle contains less than the total amount in the two small bottles. 13 Chocolate bars are produced independently in two sizes, small and large. The amount in each bar is distributed normally and independently as S » N(21, 5) and L » N(90, 15). a One of each type of bar is selected at random. Find the probability that the large bar contains more than five times the amount in the small bar. b One large and five small bars are selected at random. Find the probability that the large bar contains more than the total amount in the five small bars. 14 In a gambling game you bet on the outcomes of two spinners. These outcomes are X and Y with the following probability distributions: x

¡3

¡2

3

5

y

¡3

2

5

P(X = x)

0:25

0:25

0:25

0:25

P(Y = y)

0:5

0:3

0:2

a Briefly explain why these are well-defined probability distributions. b Find the mean and standard deviation of each random variable. c Suppose it costs $1 to get a spinner spun and you receive the dollar value of the outcome. For example, if the result is 3 you win $3, but if the result is ¡3 you need to pay an extra $3. In which game are you likely to achieve a better result? On average, do you expect to win, lose, or break even? Justify your answer.

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d Comment on the differences in standard deviation.

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e The players are now invited to play a $1 game using the sum of the scores obtained on each of the spinners. For example, if the sum of the spinners is 10, you receive $10 after paying out $1. Effectively you win $9. i Copy and complete the table below to show the probability distribution of X + Y . A grid may help you do this. x+y

¡6

¡5

P(x + y)

10

::::

0:125

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ii Calculate the mean and standard deviation of the variable U = X + Y . iii Are you likely to win, lose, or draw in the new game? Justify your answer.

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B

DISCRETE RANDOM VARIABLES

In this section we present important examples of discrete random variables and examine their cumulative distribution functions (CDF).

DISCRETE UNIFORM A discrete uniform random variable X takes n distinct values x1 , x2 , ...., xn , and the probability mass function is a constant. P(X = xi ) = X has CDF F (x) = P(X 6 x) =

P xi 6 x

1 , i = 1, 2, ...., n. n

P(X = xi ).

We write X » DU(n). For example, suppose an unbiased 6-sided die is labelled ¡2, ¡1, 0, 1, 2, 3. The random variable X for the possible outcomes is X » DU(6) since P(X = ¡2) = P(X = ¡1) = P(X = 0) = P(X = 1) = P(X = 2) = P(X = 3) = 16 . The probability that the outcome from the roll of this die is less than or equal to 1 is F (1) = P(X 6 1) = P(X = ¡2) + P(X = ¡1) + P(X = 0) + P(X = 1) = 46 .

BERNOULLI A Bernoulli random variable X has two possible values 1 (‘success’) and 0 (‘failure’). P(X = 1) = p = the probability of success in one trial P(X = 0) = 1 ¡ p = the probability of failure in one trial where p is a constant, 0 6 p 6 1. X has CDF F (x) =

x P

pk (1 ¡ p)1¡k ,

k=0

but note that there are only two values: F (0) = 1 ¡ p and F (1) = 1. We write X » B(1, p). Example 9 Consider the toss of an unbiased coin. Let X = 1 if a head is tossed and let X = 0 if a tail is the result. a Show that X is a Bernoulli variable. b Suppose that the CDF of X is F (x). Find F (0) and F (1), and interpret their meaning. a P(X = 1) =

1 2 1 2

=p

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P(X = 0) = = 1 ¡ p ¡ ¢ ) X » B 1, 12 .

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b F (0) is the probability of no heads. F (0) = P(tail) = P(X = 0) =

1 2

F (1) is the probability of at most one head. or F (1) =

F (1) = P(X 6 1) 1 2

+

1 2

pk (1 ¡ p)1¡k

k=0 0

= P(X = 0) + P(X = 1) =

1 P

= p (1 ¡ p)1 + p1 (1 ¡ p)0

=1

= (1 ¡ p) + p

=1 The probability of “no heads or one head” covers all possibilities and therefore it is a certain event.

BINOMIAL A binomial random variable X has n + 1 distinct possible values, x = 0, 1, 2, ...., n where x is the number of successes in n independent Bernoulli trials B(1, p). ¡ ¢ Thus P(X = x) = nx px (1 ¡ p)n¡x , x = 0, 1, 2, ...., n, where p is a constant, 0 6 p 6 1.

When n = 1, X is a Bernoulli random variable.

X has CDF F (x) = P(X 6 x) x ¡ ¢ P n k n¡k = k p (1 ¡ p) k=0

We write X » B(n, p). We note from the HL Core course that E(X) = np and Var(X) = np(1 ¡ p). Example 10 A bag contains 3 red balls and 4 blue balls, all identical in shape. Consider n = 8 trials of randomly choosing a ball from the bag, noting its colour, and then replacing the ball in the bag. Find the probability of choosing up to 2 red balls. Let X be the number of red balls selected in 8 such (independent) trials. ¡ ¢ ¡ ¢x ¡ 4 ¢8¡x , x = 0, 1, 2, ...., 8 and X » B(8, 37 ). Then P(X = x) = x8 37 7 The probability of choosing up to 2 red balls is F (2) =

2 ¡ ¢¡ ¢ ¡ ¢ P 8 3 k 4 8¡k k=0

=

¡ 4 ¢8 7

k

+8

7

7

¡ 3 ¢1 ¡ 4 ¢7 7

7

+

¡ 8 ¢ ¡ 3 ¢2 ¡ 4 ¢6 2

7

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EXERCISE B.1 1 The discrete random variable X is such that P(X = x) = k, for x = 5, 10, 15, 20, 25, 30. Find: a the probability distribution of X

For more questions on the uniform and binomial distributions, see the Core course Chapter 25.

b ¹, the expected value of X c P(X < ¹) d ¾, the standard deviation of X.

2 Consider the spinner alongside. Let X = 1 if the outcome is red, and X = 0 if the outcome is white. a Show that X is a Bernoulli variable. b Suppose that the CDF of X is F (x). Find F (0) and F (1), and interpret their meaning. 3 Consider the random variable X such that X » B(7, p), p < 0:5, and P(X = 4) = 0:097 24. Find P(X = 2). 4 In Statsville, the probability that it will rain on any given day in August is 0:35 . Calculate the probability that in a given week in August in Statsville, it will rain on: a exactly 3 days

b at least 3 days

c at most 3 days

d exactly 3 days in succession.

State any assumptions made in your calculations. 5 A box contains a very large number of red and blue pens. The probability that a randomly selected pen is blue, is 0:8 . How many pens would you need to select to be more than 90% certain of picking at least one red pen? State any assumptions made in your calculations. 6 A satellite relies on solar cells for its operation and will be powered provided at least one of its cells is working. Solar cells operate independently of each other, and the probability that an individual cell fails within one year is 0:7 . a For a satellite with 15 solar cells, find the probability that: i all 15 cells fail within one year ii the satellite is still operating at the end of one year. b For a satellite with n solar cells, find the probability that it is still operating at the end of one year. c Hence, find the smallest number of cells required so that the probability of the satellite still operating at the end of one year is at least 0:98 . 7 Seventy percent (70%) of the mail to ETECH Couriers is addressed to the Accounts Department. a In a batch of 20 letters, what is the probability that there will be at least 11 letters to the Accounts Department?

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b On average 70 letters arrive each day. Find the mean and standard deviation of the number of letters to the Accounts Department each day.

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STATISTICS AND PROBABILITY

8 The table gives information about the destination and type of parcels handled by ETECH Couriers.

Destination

a Find the probability that a parcel is being sent interstate given that it is priority paid. Hint: Use Bayes theorem, see HL Core text, Chapter 24.

29

Priority

Standard

Local

40%

70%

30%

Country

20%

45%

55%

Interstate

25%

70%

30%

International

15%

40%

60%

b If two standard parcels are selected, find the probability that exactly one will be leaving the state, either interstate or international. 9 At a school fete fundraiser, an unbiased spinning wheel has numbers 1 to 50 inclusive. a Find the probability of getting a multiple of 7 in one spin of the wheel. b If the wheel is spun 500 times during the day, what is the likelihood of getting a multiple of 7 more than 15% of the time? c Suppose 20 people play each time the wheel is spun. When a multiple of 7 comes up, $5 is paid to players, but when it does not the players must pay $1. i How much would the wheel be expected to make or lose for the school if it was spun 500 times? ii Find the probability that the school will lose money if the wheel is spun 500 times during the day.

GEOMETRIC Suppose a sports magazine gives away photographs of famous football players. 15 photographs are randomly placed in every 100 magazines. Let X be the number of magazines you purchase before you get a photograph. P(X = 1) = P(the first magazine contains a photo) = 0:15 P(X = 2) = P(the second magazine contains a photo) = 0:15 £ 0:85 P(X = 3) = P(the third magazine contains a photo) = 0:15 £ (0:85)2 ) P(X = x) = 0:15 £ (0:85)x¡1 for x = 1, 2, 3, 4, .... . This is an example of a geometric distribution. If X is the number of independent Bernoulli trials B(1, p), 0 < p 6 1, needed to obtain a successful outcome, then X is a geometric discrete random variable and has probability mass function P(X = x) = p(1 ¡ p)x¡1 where x = 1, 2, 3, 4, .... . The CDF is F (x) = P(X 6 x) =

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p(1 ¡ p)k¡1 for x = 1, 2, 3, 4, .... .

k=1

We write X » Geo(p).

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Example 11 In a spinning wheel game with numbers 1 to 50 on the wheel, you win if you get a multiple of 7. Assuming the game is fair, find the probability that you will win: a after exactly four games

b after no more than three games

c after more than three games. Let X be the number of games played until you win. 7 = 0:14 and 1 ¡ p = 0:86 X » Geo(p) where p = 50 a

P(X = 4) = p(1 ¡ p)3 = 0:14 £ (0:86)3 ¼ 0:0890

b

P(wins after no more than three games) = P(X 6 3) = p + p(1 ¡ p) + p(1 ¡ p)2 = p(1 + (1 ¡ p) + (1 ¡ p)2 ) = 0:14(1 + 0:86 + 0:862 ) ¼ 0:364 or

P(wins after no more than three games) = P(X 6 3) = 1 ¡ P(does not win in any of the first three games) = 1 ¡ (1 ¡ p)3 = 1 ¡ 0:863 ¼ 0:364

c P(wins after more than three games) = P(X > 3) = 1 ¡ P(X 6 3) ¼ 1 ¡ 0:364 ffrom bg ¼ 0:636 Theorem 9 Suppose X » Geo(p), 0 < p < 1, is a geometric discrete random variable. x P p(1 ¡ p)k¡1 = 1 ¡ (1 ¡ p)x 1 F (x) = P(X 6 x) = k=1

2 The modal value of X is x = 1. x P

1

Proof:

p(1 ¡ p)k¡1 = p

k=1

x P

(1 ¡ p)k¡1

k=1

= pf1 + (1 ¡ p) + (1 ¡ p)2 + :::: + (1 ¡ p)x¡1 g ½ ¾ 1 ¡ (1 ¡ p)x fsince we have a finite GSg =p p

= 1 ¡ (1 ¡ p)x 2 P(X = 1) = p P(X = x) = p(1 ¡ p)x¡1 6 p for x > 1, since 0 < p < 1.

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) P(X = 1) = p is the highest probability in the distribution. ) x = 1 is the mode of the distribution of X » Geo(p).

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NEGATIVE BINOMIAL (PASCAL’S DISTRIBUTION) If X is the number of independent Bernoulli trials B(1, p), 0 < p < 1, required for r successes then X has a negative binomial distribution. We observe that if r = 1 then the negative binomial distribution is a geometric distribution. Example 12 In grand slam mens tennis, the player who wins a match is the first player to win 3 sets. Suppose that P(Novak beats Rafael in one set) = 0:55 . Find the probability that when Novak plays Rafael in the grand slam event: a Novak wins the match in three sets c Novak wins the match in five sets

b Novak wins the match in four sets d Rafael wins the match.

Let X be the number of sets played until Novak wins. a

P(X = 3) = (0:55)3 ¼ 0:166

P(X = 4) = P(RNNN or NRNN or NNRN) = 3 £ 0:553 £ 0:451 ¼ 0:225

c

P(X = 5) = P(RRNNN or RNRNN or RNNRN or NRRNN or NRNRN or NNRRN) = 6 £ 0:553 £ 0:452 ¼ 0:202

d

P(Rafael wins the match) = 1 ¡ P(Novak wins the match) = 1 ¡ (0:553 + 3 £ 0:553 £ 0:45 + 6 £ 0:553 £ 0:452 ) ¼ 0:407

b

Examining b from the above Example 12, we notice that ¡ ¢ P(X = 4) = P(Novak wins 2 of the first 3 and wins the 4th) = 32 (0:55)2 (0:45)1 £ 0:55 | {z } binomial Generalising, P(X = x) = P(r ¡ 1 successes in x ¡ 1 independent trials and success in the last trial) ¡ ¡1 ¢ r¡1 (1 ¡ p)x¡r £ p = xr¡ 1 p ¡ x¡1 ¢ r = r¡1 p (1 ¡ p)x¡r In repeated independent Bernoulli trials, B(1, p), 0 < p < 1, where p is the probability of success in each trial, let X denote the number of trials needed to gain r successes. X has a negative binomial distribution with probability mass function ¡ ¡1 ¢ r x¡r P(X = x) = xr¡ , r > 1, where x = r, r + 1, r + 2, .... 1 p (1 ¡ p) The CDF is F (x) = P(X 6 x) =

x ¡ ¢ P k¡1 k=r

r¡1

pr (1 ¡ p)k¡r where 1 6 r 6 k 6 x.

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We write X » NB(r, p) and say X is a negative binomial random variable.

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Example 13 On each point in a badminton set, Dan has probability 0:52 of beating Chong. Suppose they play one set to 21 points. Find the probability that Dan wins the set 21-19. Let X be the number of points played until Dan wins. ) X » NB(21, 0:52) ) P(Dan wins 21-19) = P(X = 40) ¡ ¢ 21 19 = 39 20 0:52 0:48 ¼ 0:0658

EXERCISE B.2 1 X is a discrete random variable where X » Geo(0:25). Without using technology, calculate: a P(X = 4)

b P(X 6 2)

c P(X > 3)

2 Suppose X » Geo(p), 0 < p < 1. Show that the probability distribution is well defined, so 1 P P(X = i) = 1. i=1

3 In a game of ten-pin bowling, Xu has a 29% chance of getting a ‘strike’ with every bowl he attempts at all ten pins. a Find the probability of Xu getting a ‘strike’ after exactly 4 bowls. b Find the probability that Xu will take 7 bowls to score 3 ‘strikes’. 4 Suppose X » Geo(p) and that the probability that the first success is obtained on the 3rd attempt is 0:023 987. If p > 0:5, find P(X > 3). 5 In any game of squash, Paul has a 65% chance of beating Eva. To win a match in squash, a player must win three games. a State the nature of the distribution in a squash match. b Find the probability that Eva beats Paul by 3 games to 1. c Find the probability that Eva beats Paul in a match of squash. 6 From past experience, Fred has a 72% chance of driving a golf ball in excess of 230 m. Find the probability that: a Fred needs 5 drives in order to hit one in excess of 230 m. b Fred needs 12 drives in order to exceed 230 m 4 times.

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7 At a luxury ski resort in Switzerland, the probability that snow will fall on any given day in the snow season is 0:15 . a If the snow season begins on November 1st, find the probability that the first snow will fall on November 15. b Given that no snow fell during November, a tourist decides to wait no longer to book a holiday. On December 1st, the tourist decides to book for the earliest date for which the probability that the first snow will have fallen is greater than 0:85 . Find the exact date of the booking.

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STATISTICS AND PROBABILITY

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8 In a board game for four players, the players must each roll two fair dice in turn to get a difference of “no more than 3”, before they can move their marker on the board. a Find the probability of getting a difference of “no more than 3” when rolling two unbiased dice. b Player 1 rolls the dice first. Find the probability that player 1 is the first to move his counter, and this happens on his second roll.

POISSON The Poisson distribution was studied in Chapter 25 of the Core text. X is a Poisson random variable if E(X) = Var(X) = m > 0 and X has

mx e¡m , where x = 0, 1, 2, 3, .... . probability mass function P(X = x) = x!

The CDF of X is F (x) = P(X 6 x) =

x P mk e¡m

k!

k=0

We write X » Po(m).

0! = 1

for x = 0, 1, 2, 3, .... .

We have seen that a binomial random variable is used to describe the number of successes in a certain number of independent Bernoulli trials. A Poisson random variable X can be interpreted as the number of successes (or occurrences) in an interval of given, specific length, when the following conditions hold: 1 The average number E(X) = m, of occurrences is known and is constant for all intervals of the given, specific length. 2 The number of occurrences in such intervals are independent when the intervals are disjoint. 3 The probability of success in any given trial is small. Examples of random variables which can be modelled with a Poisson distribution include: ² the number of typing errors per page in a book. ² the number of telephone calls per hour received by an office. Example 14 Let X be the number of patients that arrive at a hospital emergency room per hour. Patients arrive at random and the average number of patients per hour is constant. a Explain why X is a Poisson random variable. b Suppose we know that 3Var(X) = [E(X)]2 ¡ 4. i Find the mean of X. ii Find P(X 6 4). c Let Y be the number of patients admitted to the hospital Intensive Care Unit each hour. Suppose it also has a Poisson distribution with Var(Y ) = 3, and that Y is independent of X. i Find E(X + Y ) and Var(X + Y ). ii What do you suspect about the distribution of X + Y ?

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d Let U be the random variable defined by U = X ¡ Y . i Find the mean and variance of U . ii Comment on the distribution of U .

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34

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a X is a Poisson random variable as the average number of patients arriving at random per hour is constant. We assume that the numbers of patients arriving each hour are independent. i Since E(X) = Var(X) = m, we find 3m = m2 ¡ 4 ) m2 ¡ 3m ¡ 4 = 0 ) (m ¡ 4)(m + 1) = 0 ) m = 4 or ¡1 But m > 0, so m = 4

b

ii X » Po(4). P(X 6 4) ¼ 0:629 c

E(X + Y ) = E(X) + E(Y ) =4+3 fE(Y ) = Var(Y ) = 3g =7

i

Var(X + Y ) = Var(X) + Var(Y ) =4+3 =7

ii Since E(X + Y ) = Var(X + Y ), we suspect X + Y is a Poisson random variable, and that X + Y » Po(7). d

E(U ) = E(X ¡ Y ) = E(X) ¡ E(Y ) =4¡3 =1

i

Var(U ) = Var(X ¡ Y ) = Var(X) + Var(Y ) =4+3 =7

We will prove in the next exercise that the sum of two independent Poisson random variables is itself a Poisson random variable.

ii Since E(U ) 6= Var(U ), the variable U = X ¡ Y cannot be Poisson.

EXERCISE B.3 1 X is a discrete random variable such that X » Po(¹) and P(X = 2) = P(X = 0) + 2P(X = 1). b Hence, evaluate P(1 6 X 6 5).

a Find the value of ¹.

2 Let X be the number of emergency calls made to the police per hour. The calls arrive at random, and the average number of calls per hour is constant. a Explain why X is a Poisson random variable. b Suppose we know that 2Var(X) = [E(X)]2 ¡ 15. i Find the mean of X. ii Find P(X 6 3). 3 In a mining process, the workers regularly use chains of length 50 metres. It is known that chains from a particular manufacturer have faults at the average rate of 1 per every kilometre of chain. a Find the probability that there will be: i no faults in a 50 metre length of chain ii at most two faults in the 50 metre length of chain.

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b A chain is considered ‘safe’ if there is at least a 99:5% chance there will be no more than 1 fault in 50 m of chain. Should the chains from this manufacturer be considered ‘safe’?

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STATISTICS AND PROBABILITY

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4 A receptionist in an international school receives on average five internal calls per 20 minutes and ten external calls per half hour. a Calculate the probability that the receptionist will receive exactly three calls in five minutes. b On average, how many calls will the receptionist receive every five minutes? Give your answer to the nearest integer. c Find the probability that the receptionist receives more than five calls in: i 5 minutes ii 7 minutes.

Example 15 Consider two independent Poisson random variables X and Y , both with mean m. Prove that X + Y is a Poisson random variable with mean 2m. Since X and Y are two Poisson random variables with mean m, P(X = x) = my e¡m P(Y = y) = . y!

mx e¡m and x!

Since X and Y are independent, P(X = x and Y = y) = P(X = x) £ P(Y = y) ) P(X + Y = k) =

k P

(P(X = i and Y = k ¡ i))

i=0

=

k P

P(X = i) £ P(Y = k ¡ i)

i=0

=

k P mi e¡m

i!

i=0

£

mk¡i e¡m (k ¡ i)!

=

k e¡2m P k! mi mk¡i k! i=0 i!(k ¡ i)!

=

e¡2m (2m)k k!

fBinomial theoremg

) X + Y is a Poisson random variable with mean 2m. a Let X » Po(m1 ) and Y » Po(m2 ) be two independent Poisson random variables. Prove that X + Y has probability mass function

5

(m1 + m2 )k e¡(m1 +m2 ) , k = 0, 1, 2, .... k! and thereby prove that X + Y is a Poisson random variable with distribution Po(m1 + m2 ). P(X + Y = k) =

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b Prove by induction that the sum of n independent Poisson random variables X1 , X2 , ...., Xn , with corresponding means m1 , m2 , ...., mn respectively, is a Poisson random variable with distribution Po(m1 + m2 + :::: + mn ).

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36

STATISTICS AND PROBABILITY

THE MEAN AND VARIANCE OF DISCRETE RANDOM VARIABLES We have seen that to calculate the mean and variance of a discrete random variable we use: P ² the mean E(X) = ¹ = xi pi P 2 ² the variance Var(X) = ¾ = (xi ¡ ¹)2 pi P 2 = xi pi ¡ ¹2 = E(X 2 ) ¡ fE(X)g2 Example 16 Consider X » DU(n) where X = 1, 2, ...., n, which is a special case of a discrete uniform random variable. Given that 12 + 22 + 32 + :::: + n2 = n2 ¡ 1 . 12

and Var(X) = E(X) =

n(n + 1)(2n + 1) n+1 for all n in Z + , show that E(X) = 6 2

P

xi pi ³ ´ ³ ´ ³ ´ ³ ´ 1 1 1 1 =1 +2 +3 + :::: + n n n n 1 = (1 + 2 + 3 + 4 + :::: + n) n

=

h

1 n

n

where 1 + 2 + 3 + :::: + n is an arithmetic series with u1 = 1 and d = 1

i

n (2u1 + (n ¡ 1)d) 2

= 12 [2 + (n ¡ 1)] =

n+1 2

P

xi2 pi ¡ ¹2 ³ ´ ³ ´ ³ ´ ³ ´ ³ ´ 1 1 1 1 n+1 2 = 12 + 22 + 32 + :::: + n2 ¡

Var(X) =

n

n

n

2

n

1 (n + 1)2 = (12 + 22 + 32 + :::: + n2 ) ¡ n 4

h

i

n(n + 1)(2n + 1) (n + 1)2 ¡ 6 4

=

1 n

=

(n + 1)(2n + 1) (n + 1)2 ¡ 6 4

h

i

2n + 1 n+1 ¡ 6 4 h i 4n + 2 3n + 3 = (n + 1) ¡ 12 12 h i n¡1 = (n + 1) 12

= (n + 1)

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n2 ¡ 1 12

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STATISTICS AND PROBABILITY

For X » DU(n) in general, X takes values x1 , x2 , ...., xn where P(X = xi ) = pi = for all i = 1, ...., n. The mean and variance are calculated as: E(X) =

n P

xi pi =

i=1

Var(X) =

n P i=1

37

1 n

n 1 P xi n i=1

xi2 pi ¡ fE(X)g2

½

n 1 P 1 = x2¡ 2 n i=1 i n

n P

¾2 xi

.

i=1

By similar direct calculation of the mean and variance for discrete distributions, the results in the following table can be obtained. We will return to them later in the course when we study probability generating functions. Distribution

Notation

Probability mass function

Mean

Variance

Bernoulli

X » B(1, p)

px (1 ¡ p)1¡x for x = 0, 1

p

p(1 ¡ p)

np

np(1 ¡ p)

m

m

1 p

q p2

for x = r, r + 1, ....

r p

rq p2

1 for x = 1, ...., n n

n+1 2

n2 ¡ 1 12

¡n¢ Binomial

X » B(n, p)

Poisson

X » Po(m)

Geometric

X » Geo(p)

X » NB(r, p)

Discrete uniform (special case)

X » DU(n)

a Prove that x

¡n¢ x

=n

¡ n¡1 ¢ x¡1

px (1 ¡ p)n¡x

for x = 0, 1, ...., n mx e¡m for x = 0, 1, .... x!

pq x¡1 where q = 1 ¡ p, for x = 1, 2, ....

¡ x¡1 ¢

Negative binomial (Pascal’s)

Example 17

x

r¡1

pr qx¡r where q = 1 ¡ p,

.

b Hence prove that for a Binomial random variable, the mean is equal to np. a Proof:

LHS = x

¡n¢

RHS = n

x

n! =x£ (n ¡ x)!x!

=

¡ n¡1 ¢ x¡1

=n£

n! (n ¡ x)!(x ¡ 1)!

=

(n ¡ 1)! (n ¡ x)!(x ¡ 1)!

n! (n ¡ x)!(x ¡ 1)!

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) LHS = RHS as required

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38

STATISTICS AND PROBABILITY

b If X » B(n, p) then P (x) = n P xP (x) ) ¹ = E(X) = = = =

x=0 n P x=0 n P x=1 n P

x

¡n¢

x

¡n¢

n

¡ n¡1 ¢

x

x

= np = np

r=0

= np

nP ¡1 ¡ r=0

where q = 1 ¡ p.

n¡1 r n¡1 r

fsince when x = 0, the term is 0g

px qn¡x

x¡1

nP ¡1 ¡

px q n¡x

px q n¡x

n ¡ ¢ P n¡1

x=1

x

px q n¡x

x¡1

x=1

¡n¢

¢ ¢

fusing the above resultg

px¡1 q n¡x pr qn¡(r+1)

freplacing x ¡ 1 by rg

pr q(n¡1)¡r

= np(p + q)n¡1

fBinomial theoremg

= np £ 1 = np

THE POISSON APPROXIMATION TO THE BINOMIAL DISTRIBUTION In the following example we observe how, in certain cases, a binomial random variable can be approximated by a Poisson random variable. Example 18 Sheep are transported to the city using big trucks which carry 500 sheep at a time. On average, 0:8% of the sheep have to be removed on arrival because of illness. a Describe the nature of the random variable X, which indicates the number of ill sheep on arrival. b State the mean and variance of X. c Find the probability that on a truck carrying 500 sheep: i exactly three are ill on arrival ii at least four are ill on arrival. d By inspection of your answer to b, comment as to what other type of random variable X may approximate. e Use the approximation from d to repeat the calculations in c. Hence verify the validity of the approximation. a X is a binomial random variable and X » B(500, 0:008) ¾ 2 = np(1 ¡ p) = 4 £ 0:992 ¼ 3:97

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i P(X = 3) ¼ 0:196

c

75

b ¹ = np = 500 £ 0:008 = 4

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STATISTICS AND PROBABILITY

39

d Using b, ¹ ¼ ¾ 2 which suggests we may approximate X using a Poisson distribution. In particular, X is approximately distributed as Po(4). P(X = 3) ¼ 0:195 X

e

P(X > 4) = 1 ¡ P(X 6 3) ¼ 0:567 X

These results are excellent approximations to those in c.

The previous example prompts the question: Under what conditions can a binomial random variable be approximated by a Poisson random variable? Let X » B(n, p) be a binomial random variable. X has probability mass function P (x) = P(X = x) = and E(X) = np.

¡n¢ x

px (1 ¡ p)n¡x where x = 0, 1, 2, ...., n

Suppose np = m, a constant. ) p=

m n

For m, x fixed constants: P (x) =

´ ¡ n ¢ ³ m ´x ³ m n¡x 1 ¡ x n

n

³ ´ m n¡x 1¡ n n³ ´ ³ ´³ ´ ³ ´o ³ ´ x m n n¡1 n¡2 n¡x+1 m n¡x = :::: 1¡ x! n n n n n n ³ ´³ ´ ³ ´o ³ ´ ³ ´ x m 1 2 (x ¡ 1) m n m ¡x = 1 1¡ 1¡ :::: 1 ¡ 1¡ 1¡ n(n ¡ 1)(n ¡ 2) :::: (n ¡ x + 1) mx = x! nx

x!

n

n

n

n

n

For a Poisson distribution, the probability p of success in a given trial is small. If we consider p ! 0, we require n ! 1 in order to keep np = m constant. n ³ ´³ ´ ³ ´o 1 2 (x ¡ 1) 1 1¡ 1¡ :::: 1 ¡ ! 1, ) n n n ³ ´ m ¡x 1¡ ! 1, n ³ ´n ¡m and 1 + ! e¡m frefer to Calculus Optiong n

) P (x) !

mx e¡m x!

Thus the binomial distribution approaches a Poisson distribution when p ! 0 and n ! 1. For X » B(n, p) a binomial random variable, X can be approximated by the Poisson random variable Y » Po(np) provided n is large enough and p is small enough. In general, Y is a reasonable approximation provided n > 50 and p 6 0:1 .

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Where appropriate in the following exercise, clearly state the type of discrete distribution used as well as answering the question.

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40

STATISTICS AND PROBABILITY

EXERCISE B.4 1 Suppose X » Geo(0:333). Find: a the mean of X

b the standard deviation of X.

2 In a game of ten-pin bowling, Win has a 25% chance of a strike in each frame. a Find the average number of bowls Win requires to get a strike. b What is the average number of bowls Win will need to get two strikes? 3 A dart player has a 5% chance of hitting the bullseye with every attempt. Find: a the expected number of throws for this player to get a bullseye b the standard deviation in throws for this player to get a bullseye. 4 A spinning wheel has the numbers 1 to 40 inclusive on it. Assuming that the wheel is unbiased, find the mean and standard deviation of all the possible scores when the wheel is spun. 5 In an average working week, an office confectionary dispenser breaks down six times. Assume the working week is Monday to Saturday, with each day including the same number of working hours. Which of the following is most likely to occur? A The machine breaks down three times during the week. B The machine breaks down once on Saturday. C The machine breaks down less than seventeen times in a 4 week period. 6

In a World Series contest between the Redsox and the Yankees, the first team to win four games is declared world champion. Recent evidence suggests that the Redsox have a 53% chance of beating the Yankees in any game. Find the probability that: a the Yankees will beat the Redsox in exactly five games b the Yankees will beat the Redsox in exactly seven games c the Redsox will be declared world champions. d On average, how many games would it take for the Redsox to win four games against the Yankees? Comment on your result.

7 During the busiest period on the internet, you have a 62% chance of connecting to a government website. If you do not get through, you keep trying until you do connect. Let X be the number of times you have to try in order to get through. a Stating any necessary assumptions, identify the nature of the random variable X. b Find P(X > 3). c Find the mean and standard deviation of the random variable X. 8 A large aeroplane has 250 passenger seats. From years of business experience, the airline has found that on average 3:75% of travellers who have bought tickets do not arrive for any given flight. The airline sells 255 tickets for this large aeroplane on a particular flight. Let X be the number of ticket holders who do not arrive for the flight. a State the distribution of X. b Calculate the probability that more than 250 ticket holders will arrive for the flight. c Calculate the probability that there will be empty seats on this flight.

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d For the variable X, calculate the:

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STATISTICS AND PROBABILITY

41

e Hence use a suitable approximation for X to calculate the probability that: i more than 250 ticket holders will arrive for the flight ii there will be empty seats on this flight. f Use your answers to determine whether the approximation was a good one. 9 It costs E15 to play a game where you have to randomly select a marble from ten differently marked marbles in a barrel. The marbles are marked 10 cents, 20 cents, 30 cents, 40 cents, 50 cents, 60 cents, 70 cents, E15, E30, and E100, and you receive the marked amount in return for playing the game. a Define a random variable X which is the outcome of selecting a marble from the barrel. b Find E(X) and Var(X). c Briefly explain why you cannot use the rules given for DU(n) to find the answers to b above. d The people who run the game expect to make a profit but want to encourage people to play by not charging too much. i Find, to the nearest 10 cents, the smallest amount they need to charge to still expect to make a profit. ii Find the expected return to the organisers if they charge E16 per game, and a total of 1000 games are played in one day. 10 A person raising funds for cancer research telephones people at random asking for a donation. From past experience, he has a 1 in 8 chance of being successful. a Describe the random variable X that indicates the number of calls made before he is successful and someone makes a donation. State one assumption made in your answer. b Find the average number of calls required for success, and the standard deviation of the number of calls for success. c Find the probability that it takes less than five calls to be successful. 11 The probability that I dial a wrong number is 0:005 when I make a telephone call. In a typical week I will make 75 telephone calls. a Describe the distribution of the random variable T that indicates the number of times I dial a wrong number in a week. b In a given week, find the probability that: i I dial no wrong numbers, P(T = 0) ii I dial more than two wrong numbers, P(T > 2). c Find E(T ) and Var(T ). Comment on your results. d By approximating T with a Poisson distribution, again find the probability that in a given week: i I dial no wrong numbers ii I dial more than two wrong numbers. e Discuss your results in b and d. a Consider 1 + q + q 2 + q 3 + :::: where 0 < q < 1. i Find the sum to infinity of this series.

12

ii Hence, show that 1 + 2q + 3q 2 + 4q3 + :::: =

1 P x=1

xqx¡1 =

1 (1 ¡ q)2

for 0 < q < 1.

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1 b If X » Geo(p), prove that E(X) = . p

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42

STATISTICS AND PROBABILITY

C

CONTINUOUS RANDOM VARIABLES

In this section we present important examples of continuous random variables and examine their cumulative distribution functions (CDF). Refer to Chapter 26 of the Core text to revise the properties of continuous random variables. In Section A we saw that a continuous random variable X with probability distribution function f (x) with domain [a, b], has cumulative distribution function F (x), where F (x) = P(X 6 x) Z x = f (t) dt

y y = f(t)

Area = F(x)

a

= area under the curve y = f(t) between t = a and t = x. a

x

b

t

Depending on the form of the function f (t), this area can sometimes be found using simple methods, for example, by finding the area of a triangle or rectangle. Example 19 The continuous random variable X has PDF f (x) = kx, 0 6 x 6 6, where k is a constant. Find: a k b the tenth percentile of the random variable X.

Z a

6

Since

y

0

Z

f(x) dx = 1, 6

kx dx = 1 · ¸6 x2 =1 ) k

f(x) = kx

0

2

We could use the area of a triangle formula instead of integrating.

0

) k(18 ¡ 0) = 1 1 ) k = 18 b We need to find c such that P(X < c) = 0:10 6

c

1 2

)

£c£

x

c = 0:1 18

fThe area of the shaded triangleg

) c2 = 3:6 ) c ¼ 1:90 fas c > 0g The 10th percentile ¼ 1:90

THE MEAN AND VARIANCE OF A CONTINUOUS RANDOM VARIABLE

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From Section A we have the following formulae for calculating the mean and variance of a continuous random variable X: Z ² E(X) = ¹ = x f (x)dx Z ² Var(X) = ¾2 = (x ¡ ¹)2 f (x)dx = E((X ¡ ¹)2 ) Z x2 f (x)dx ¡ ¹2 = E(X 2 ) ¡ ¹2 =

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43

STATISTICS AND PROBABILITY

CONTINUOUS UNIFORM A continuous uniform random variable X has: ² domain [a, b], where a, b are constants. The possible values for X are all x such that a 6 x 6 b. ² PDF f (x) =

1 for a 6 x 6 b. b¡a

y

f (x) =

1 b¡a

Its graph a horizontal line segment as shown.

a

Z

x

² CDF F (x) = P(X 6 x) = =

Zax a

=

1 b¡a

f (x) =

y

f (t) dt

1 b¡a

1 dt b¡a

x

b 1 b¡a

Area = F(x)

x¡a b¡a

a

x

b

= the area of the rectangle shown. We write X » U(a, b). Example 20 If X » U(a, b) is a continuous uniform random variable, show that: a ¹=

a+b 2

b Var(X) =

Since X » U(a, b), its PDF is f(x) = a ¹ = E(X) Z b x = a

= =

b¡a

1 b¡a b2 2

¡

·

x2 2

dx

a

¸b = a

a2 2

=

b¡a

b¡a

1 b¡a b3 3

¡

·

x3 3

a3 3

b¡a

2

¸b

¡

b3 ¡ a3 = ¡ 3(b ¡ a)

¡ a

³ ³

³

´ a+b 2 2

´ a+b 2 2

´ a+b 2 2

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(b ¡ a)2 12

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b2 ¡ 2ab + a2 12

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=

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4b2 + 4ab + 4a2 3a2 + 6ab + 3b2 ¡ 12 12

100

=

50

a+b 2

75

=

25

(b ¡ a)(b2 + ab + a2 ) a2 + 2ab + b2 ¡ 3(b ¡ a) 4

0

=

5

(b + a)(b ¡ a) 2(b ¡ a)

95

=

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1 , a 6 x 6 b. b¡a

b ¾ 2 = Var(X) = E(X 2 ) ¡ ¹2 Z b 2 ³ ´ x a+b 2 = dx ¡

b2 ¡ a2 = 2(b ¡ a)

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t

44

STATISTICS AND PROBABILITY

Example 21 The error, in seconds, made by an amateur timekeeper at an athletics meeting may be modelled by the random variable X with probability density function ½ 2:5 ¡0:1 6 x 6 0:3 f(x) = 0 otherwise. Find the probability that: b the magnitude of an error exceeds 0:1 seconds

a an error is positive

c the magnitude of an error is less than 0:2 seconds. We have X » U(¡0:1, 0:3) with PDF f(x) = 2:5 on ¡0:1 6 x 6 0:3 . a

P(X > 0) = P(0 < X < 0:3) = 0:3 £ 2:5 fArea of the given rectangleg = 0:75

b

y y = f(x)

2.5

P(magnitude > 0:1) = P(jXj > 0:1) = P(X > 0:1 or X < ¡0:1) = P(X > 0:1) = 0:2 £ 2:5 fArea of rectangleg = 0:5 y y = f(x)

2.5 -0.1

x

0.3

-0.1

c P(magnitude < 0:2) = P(jXj < 0:2) = P(¡0:2 < X < 0:2) = P(¡0:1 < X < 0:2) = (0:2 ¡ (¡0:1)) £ 2:5 = 0:75

0.3

0.1

x

y 2.5

-0.1

y = f(x)

0.2 0.3

x

EXPONENTIAL A continuous exponential random variable X has: ² domain [ 0, 1 [ . The possible values for X are all x > 0. ² PDF f (x) = ¸e¡¸x for x > 0, where ¸ > 0 is a constant. ¸

y f (x) = ¸e¡¸x

x

Z

x

¸e¡¸t dt £ ¤x = ¡e¡¸t 0

² CDF F (x) = P(X 6 x) =

0

= 1 ¡ e¡¸x .

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We write X » Exp(¸).

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45

We note that: ² ¸ > 0 is required since f (x) > 0 is necessary and e¡¸x > 0 for all x. ² f(x) is decreasing for all x > 0. This is observed from the graph y = f (x) or algebraically since f 0 (x) = ¡¸2 e¡¸x < 0 for all x > 0. Z 1 Z 1 f (x) dx = 1. Hence ¸e¡¸t dt = 1. ² Since f(x) is a PDF, 0

0

² From the graph of y = f (x) we observe that as x ! 1, f(x) ! 0 from above. ² Most of the area under the graph of y = f(x) corresponds to low values of x. Theorem 10 For a continuous exponential random variable, X » Exp(¸), E(X) =

1 1 and Var(X) = 2 . ¸ ¸

We will prove this theorem in Exercise C, question 10. Example 22 The continuous random variable X has probability density function f (x) = 2e¡2x , x > 0. a Show that f (x) is a well-defined PDF. b Find E(X) and Var(X). c Find the median and modal values of X.

Z

1

a Now 0

Z f(x) dx =

1

0

2e¡2x dx · ¸t

f (x) = 2e¡2x > 0 since e¡2x > 0 for all x 2 [ 0, 1 [ .

and

2e¡2x t!1 ¡2 0

= lim

£ ¤t = lim ¡e¡2x 0 t!1 ¡ ¢ = lim 1 ¡ e¡2t t!1

=1¡0 =1 ) f(x) is a well-defined PDF. b Since X » Exp(2), E(X) =

1 1 = 12 and Var(X) = 2 = 14 . ¸ ¸

Z

c If the median is m, we need to find m such that h³ ´ im 1 2e¡2x = 0:5 ) ¡2 0 £ ¤m ) ¡e¡2x 0 = 0:5

0

m

2e¡2x dx = 0:5 y 2

y = f(x)

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) ¡e¡2m ¡ (¡1) = 0:5 ) e¡2m = 0:5 freciprocalsg ) e2m = 2 ) 2m = ln 2 ) m = 12 ln 2 ¼ 0:347 ) the median ¼ 0:347 The mode is the value of x which gives the maximum value of f(x), ) the mode = 0.

x

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STATISTICS AND PROBABILITY

Example 23 a Find the 80th percentile of the continuous exponential random variable X with PDF f(x) = ¸e¡¸x , x > 0, giving your answer in terms of ¸. b If ¸ > 4, find possible values for the 80th percentile. Comment on your answer.

) the 80th percentile
4, 80% of the scores are less than 0:402 . 80% of the area under the curve lies in [0, 0:402], which is a very small interval compared with [ 0, 1 [.

0 £ ¡¸c ¤ 0 ¡ e ¡ e = 0:8

)

1 1 < ¸ 4

b If ¸ > 4,

a We need find c such that Z to c ¸e¡¸t dt = 0:80 0 Z c e¡¸t dt = 0:8 ) ¸ 0 · ¸c e¡¸t ) ¸ = 0:8

0.402

x

Suppose we are given the CDF of a continuous random variable. We can find its PDF using the Fundamental Theorem of Calculus, since the PDF f (x) will also be a continuous function. In particular:

Z If the CDF is F (x) =

x

f (t) dt then its PDF is given by f(x) = F 0 (x).

a

Example 24

Z Find the PDF of the random variable with CDF F (x) =

f(x) = F 0 (x) = =

d dx d dx

Z ·

x

0

x

¸e¡¸t dt, x > 0

As expected from the Fundamental Theorem of Calculus, f(x) is simply the integrand function.

¸x ¸e¡¸t ¡¸

¸e¡¸t dt.

0

0

d £ ¡¸t ¤ x = ¡e 0 dx d = (¡e¡¸x ¡ (¡1)) dx

= ¡e¡¸x (¡¸) + 0

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) the PDF is f (x) = ¸e¡¸x , x > 0

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NORMAL A continuous normal random variable X has: ² domain R , so X may take any real value x 2 R . ¡ ¢ 1 x¡¹ 2 1 ¡2 ¾ ² PDF f (x) = p e for x 2 R ,

y = f(x)

¾ 2¼

where ¹, ¾ are constants and ¾ > 0.

X

¹-¾ ¹ ¹+¾

² CDF F (x) = P(X 6 x) Z x f (t) dt =

Area = F(x) y = f(t)

¡1

= area under y = f (t) on ] ¡1, x ]. t

x

We write X » N(¹, ¾ 2 ). We note that: ² X has mean E(X) = ¹ and variance Var(X) = ¾ 2 . ² The normal curve is bell-shaped with the percentages within its portions as shown. Refer to the Core text Chapter 26 for more information.

34.13% 34.13%

0.13% 2.15%

13.59% ¹-2¾ ¹-¾

¹-3¾

² Z=

2.15%

13.59% ¹+¾ ¹+2¾

¹

0.13%

¹+3¾

X

X ¡¹ is the standard normal random variable, and Z » N(0, 1) ¾

This transformation is useful when determining an unknown mean or standard deviation. Conversion to Z-scores is also very important for understanding the theory behind confidence intervals and hypothesis testing which are dealt with later in this topic.

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X » Exp(¸)

¸e¡¸x , x > 0

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Normal

X » N(¹, ¾ 2 )

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SUMMARY OF CONTINUOUS DISTRIBUTIONS

¾

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STATISTICS AND PROBABILITY

Example 25 Consider the random variable X » N(¹, ¾ 2 ). Find the mean and standard deviation given that area A = 0:115 06 and area B = 0:135 66 .

Area = B

Area = A 13

)

¾ 36 ¡ ¹ ¼ 1:1 ¾

)

¾ 13 ¡ ¹ ¼ ¡1:2 ¾

)

X

36

Also, P(X > 36) = 0:135 66 ³ ´ 36 ¡ ¹ ) P Z> = 0:135 66

P(X < 13) = 0:115 06 ´ X ¡¹ 13 ¡ ¹ < = 0:115 06 P ¾ ¾ ³ ´ 13 ¡ ¹ ) P Z< = 0:115 06

³

)

¹

)

¹ + 1:1¾ = 36 .... (2)

¹ ¡ 1:2¾ = 13 .... (1)

(2) ¡ (1) gives 2:3¾ = 23 )

¾ = 10

Substituting in (1) gives ¹ ¡ 12 = 13 )

¹ = 25

) the mean is 25 and the standard deviation is 10.

THE NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION Suppose X » B(n, p) is a binomial random variable. For sufficiently large n, we can approximate X (discrete) by Xc (continuous) where Xc » N(np, npq) and q = 1 ¡ p.

We will prove this result later in the topic.

A useful rule to follow is: If np > 5 and nq > 5, then any value of Xc is a reasonable approximation for a value of X, provided we allow for a correction of continuity (see below).

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binomial distribution with large n.

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This can be observed by drawing column graphs for binomial distributions for different values of n and p. When n, p, and q satisfy the above, the column graph has an approximate bell-shape like the PDF of a normal distribution. The greater the values of np and nq, the closer this approximates the graph of the normal distribution.

continuous normal random distribution

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CORRECTION FOR CONTINUITY For Xc a continuous random variable, P(Xc = a) = 0 for a any constant, but for a an integer we define P(Xc = a) = P(a ¡ 0:5 6 Xc < a + 0:5), due to rounding on the real number line. Thus when approximating the discrete binomial random variable X » B(n, p) with the continuous normal random variable Xc » N(np, np(1 ¡ p)) we approximate P(X = 7) with P(6:5 6 Xc < 7:5). discrete

continuous

Also, for X 6 7 we use Xc < 7:5, and for X > 7 we use Xc > 6:5 . This is called a correction for continuity. Example 26 Consider the binomial random variable X » B(15, 0:4). a Find:

i E(X)

ii Var(X).

b Find:

i P(X 6 7)

ii P(3 6 X 6 12).

c Approximate X with an appropriate continuous normal random variable Xc . d Find: i P(Xc 6 7) ii P(3 6 Xc 6 12). Compare your answers with b. e Find: i P(Xc < 7:5) ii P(2:5 6 Xc < 12:5). Again, compare your answers with b. f Which of the approximations c or d are better? Explain your answer. a

i )

E(X) = ¹ = np E(X) = 15 £ 0:4 =6

ii )

i P(X 6 7) ¼ 0:787

b

ii

Var(X) = ¾ 2 = npq Var(X) = 6 £ 0:6 = 3:6

P(3 6 X 6 12) = P(X 6 12) ¡ P(X 6 2) ¼ 0:973

c Using a normal approximation, X is approximately distributed by Xc » N(6, 3:6). i P(Xc 6 7) ¼ 0:701 ii P(3 6 Xc 6 12) ¼ 0:942 These answers are not really close to those in b.

d

e Using the normal approximation Xc » N(6, 3:6) i P(Xc < 7:5) ¼ 0:785 ii P(2:5 6 Xc < 12:5) ¼ 0:967 These results are very close to the actual values.

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f The approximations in e are much closer. There has been a correction for continuity because the binomial distribution is discrete and the normal distribution is continuous.

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STATISTICS AND PROBABILITY

EXERCISE C Where appropriate in the following exercise, clearly state the type of discrete or continuous distribution used as well as answering the question. 1 The continuous random variable T has the probability density function ½ 1 ¡¼ 6 t 6 ¼ f(t) = 2¼ Find the mean and standard deviation of T . 0 otherwise. 2 The Australian football Grand Final is held annually on the last Saturday in September. With approximately 100 000 in attendance each year, ticket sales are heavily in demand upon release. Let X be the random variable which gives the time (in hours) required for a successful purchase of a Grand Final ticket after their release. The median value of X is 10 hours. a Give reasons why X could best be modelled by a continuous exponential random variable. b Find the value of ¸ in the PDF for the exponential random variable X. c Hence, find the probability of a Grand Final ticket being purchased after 3 or more days. d Find the average time before a Grand Final ticket is purchased. 3 Find the mean and standard deviation of a normal random variable X, given that P(X > 13) = 0:4529 and P(X > 28) = 0:1573. 8 x k. Find: a the value of k b the mean and standard deviation of the distribution. 5 It is known that 41% of a population support the Environment Party. A random sample of 180 people is selected from the population. Suppose X is the random variable giving the number in the sample who support the Environment Party. a State the distribution of X. b Find E(X) and Var(X). c Find P(X > 58). d State a suitable approximation for the random variable X and use it to recalculate c. Comment on your answer. 6 When typing a document, trainee typists make on average 2:5 mistakes per page. The mistakes on any one page are made independently of any other page. Suppose X represents the number of mistakes made on one page, and Y represents the number of mistakes made in a 52-page document. a State the distributions of X and Y . You may assume that the sum of n independent Poisson random variables, each with mean m, is itself a Poisson random variable with mean mn. b Rana is a trainee typist. Find the probability that Rana will make: i more than 2 mistakes on a randomly chosen page ii more than 104 mistakes in a 52-page document. c Now assume that X and Y can be approximated by normal random variables with the same means and variances as found above. Use the normal approximations to estimate the probabilities in b. Comment on your answers. 7 The continuous random variable X has PDF f (x) =

2 5

where 1 6 x 6 k.

a Find the value of k, and state the distribution of X.

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b Find P(1:7 6 X 6 3:2).

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8 The continuous random variable X is uniformly distributed over the interval a < x < b. The 30th percentile is 3 and the 90th percentile is 12. Find: a the values of a and b c P(5 < X < 9)

b the PDF of X d the CDF of X.

a If the random variable T » N(7, 36), find P(jT ¡ 6j < 2:3).

9

b Four random observations of T are made. Find the probability that exactly two of the observations will lie in the interval jT ¡ 6j < 2:3 . 1

¡2x

10 A continuous random variable X has PDF f (x) = 12 e

for x > 0.

a Show that X » Exp(0:5). b Hence find: i ¹X

ii ¾ X

iii the median of X

iv the 90th percentile of X.

c Use the CDF for the exponential variable to find, correct to 4 decimal places: i P(X 6 1) ii P(0:4 6 X 6 2) 11 The exponential probability density function of random variable X is defined as f (x) = ae¡ax for a > 0 and x 2 [0, 1[ . a On the same set of axes, graph y = f(x) for a = 1, 2, and 3.

Here we prove Theorem 10.

b Prove that f (x) is a well defined PDF. c Use integration by parts to show that: µ ¶ Z 1 ¡ax ¡ax i axe dx = ¡e x+ + constant a µ ¶ Z 2 2x + 2 + constant ii ax2 e¡ax dx = ¡e¡ax x2 + a a d Show that the mean and variance of the negative exponential variable X are

1 1 and 2 a a

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STATISTICS AND PROBABILITY

D

PROBABILITY GENERATING FUNCTIONS

In this section we consider discrete random variables which take values in N = f0, 1, 2, ....g. First we state some key results required for this section: 1 Finite geometric series (GS) n P

For x 2 R , x 6= 1, and n 2 N ,

xi = 1 + x + x2 + :::: + xn =

i=0

1 ¡ xn+1 . 1¡x

2 Sum of an infinite geometric series (GS) 1 P xi = 1 + x + x2 + x3 + :::: is finite (or convergent) if and only if jxj < 1. The infinite sum i=0

1 P

xi =

i=0

1 if and only if jxj < 1. 1¡x

3 Binomial formula For real constants x, y and for n 2 N : n ¡ ¢ ¡ ¢ ¡ ¢ ¡ n ¢ n¡1 P n i n¡i = yn + n1 xyn¡1 + n2 x2 y n¡2 + :::: + n¡ y + xn (x + y)n = i x y 1 x i=0

4 Exponential series 1 k P x x2 x3 =1+x+ + + :::: for all x 2 R ex = k=0

5

lim

n!1

2!

k!

fResult from the Calculus Optiong

3!

³ ´ a n 1+ = ea for all a 2 R

fResult from HL Coreg

n

6 Binomial series for r 2 Z + and jxj < 1 1 P 1 (¡r)(¡r ¡ 1) :::: (¡r ¡ (i ¡ 1)) xi (¡1)i r = (1 ¡ x)

fResult from the Calculus Optiong

i!

i=0

7 Summation identities n P n(n + 1) i = 1 + 2 + 3 + :::: + n = i=1 n P

2

i2 = 12 + 22 + 32 + :::: + n2 =

i=1

n(n + 1)(2n + 1) 6

PROBABILITY GENERATING FUNCTIONS Let X be a discrete random variable which takes values in N = f0, 1, 2, 3, ....g, and such that P(X = k) = pk , for k 2 N . The probability generating function (PGF), G(t), for X is 1 P G(t) = E(tX ) = pk tk k=0

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STATISTICS AND PROBABILITY

We note that: ² 0 6 pk 6 1 and

1 P

53

pk = 1 by the definition of the (well-defined) probability mass function for

k=0

a discrete random variable X. ² The PGF G(t) is either a finite series or an infinite series. I I

If G(t) is a finite series then G(t) is defined for all t 2 R . If G(t) is an infinite series, then it is a power series and is therefore finite (or convergent) only for t in the interval of convergence of the power series.

² A PGF G(t) defines a discrete random variable X (and its probability distribution) uniquely. Conversely, if X is a discrete random variable which takes values in N , then its PGF G(t) is unique. Example 27 a Let X be the discrete random variable which takes values 1, 2, 3, and 6, each with probability 14 . Find the PGF for X. ¡ ¢ b Let X » B 1, 16 be the Bernoulli random variable equal to the number of ‘6’s obtained when an unbiased 6-sided die is rolled once. Find the PGF for X. ¡ ¢ c Let X » Geo 16 be the geometric random variable equal to the number of rolls of an unbiased 6-sided die required to roll a ‘6’. Find the PGF for X. a G(t) = p1 t1 + p2 t2 + p3 t3 + p6 t6 = 14 (t + t2 + t3 + t6 )

Since G(t) is a finite series, G(t) is finite and therefore defined for all t 2 R .

b P(X = 0) = p0 = P(X = 1) = p1 =

5 6 1 6

P(X = k) = 0 for integers k > 2. ) the PGF for X is G(t) = p0 + p1 t + p2 t2 + :::: + 16 t for t 2 R . ¡ ¢k¡1 c X takes values 1, 2, 3, .... and P(X = k) = 16 1 ¡ 16 ¡ ¢k¡1 = 16 56 1 ¡ ¢ P 1 5 k¡1 k ) the PGF for X is G(t) = t 6 6

5 6

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if and only if jtj
0:5) ¼ 0:000 054 0 ¡ ¢ £0:57 b pb » N 0:43, 0:43100 ) P(pb > 0:62) ¼ 0:000 062 1

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This is so unlikely that we would question the accuracy of the claim that Ms Burford only gained 43% of the vote.

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STATISTICS AND PROBABILITY

81

EXERCISE E.3 1 An egg producer claims that less than 4% of the eggs delivered to a supermarket will be broken. On a busy day, 1000 eggs are delivered to this supermarket and 7% are broken. a Find the probability that this will happen.

b Briefly comment on the producer’s claim.

2 Two sevenths of households in a country town are known to own computers. Find the probability that in a random sample of 100 households, no more than 29 households own a computer. 3 A pre-election poll is run to determine the proportion of voters who favour the Labour Party (LP). The poll is based on one random sample of 2500 voters. Let p be the true proportion of voters who support the LP, and suppose p = 0:465 . a For the sampling distribution of proportions, find: i the mean ii the standard deviation. b State the sampling distribution and its normal approximation. c Hence, find the probability that: i the sample mean is less than 0:46 ii between 45% and 47% of voters in the sample favour the LP iii the proportion pb of LP supporters in the sample differs by more than 0:035 from p. d Interpret your answer in c iii. 4 Eighty five percent of the plum trees grown in a particular area produce more than 700 plums. a State the sampling distribution for the proportion of plum trees that produce more than 700 plums in this area, assuming a sample of size n. b State the conditions under which the sampling distribution can be approximated by the normal distribution. c If a random sample of 200 plum trees is selected, find the probability that: i less than 75% produce more than 700 plums ii between 75% and 87% produce more than 700 plums. d In a random sample of 500 plum trees, 350 produced more than 700 plums. i Find the probability of 350 or fewer trees producing more than 700 plums. ii Comment, giving two reasons, why this sample is possible. 5 A regular pentagon has sectors numbered 1, 1, 2, 3, 4. Find the probability that, when the pentagon is spun 400 times, the result of a 1 occurs: a more than 150 times

b less than 175 times.

6 A tyre company in Moscow claims that at least 90% of the tyres they sell will last at least 30 000 km. A sample of 250 tyres were tested, and it was found that 200 of the tyres did not last for at least 30 000 km. a State the distribution of the sample proportions, with any assumptions made.

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F

POINT ESTIMATION (UNBIASED ESTIMATORS AND ESTIMATES)

In the previous section we considered taking a sample of n independent observations (or values) from a random variable X with mean E(X) = ¹ and variance Var(X) = ¾ 2 . Let X be a random variable with mean ¹ and variance ¾ 2 . A sample of size n is taken from the population of X with replacement so the values are independent. The independent values x1 , x2 , :..., xn can be interpreted as x1 2 X1 , x2 2 X2 , ...., xn 2 Xn where Xi , i = 1, 2, ...., n are n copies of the distribution of X. E(X1 ) = E(X2 ) = :::: = E(Xn ) = ¹

Thus

and Var(X1 ) = Var(X2 ) = :::: = Var(Xn ) = ¾ 2 . It is often impractical to work with an entire population, and often the population parameters ¹, ¾ 2 , and proportion p are unknown anyway. In these cases, we work with a sample, and we use the sample statistics to estimate the population parameters. For a large enough sample, we expect: ² the sample mean x to be close in value to the population mean ¹ ² the sample proportion pb to be close in value to the population proportion p. An estimator T is a statistic, which is a function of the values in a sample, used to estimate a population parameter µ. An estimate t is a specific value of T calculated from a particular sample. For example, the number of heads obtained when an unbiased coin is tossed once, is described by ¡ ¢ X » B 1, 12 , where E(X) = ¹ = 12 and Var(X) = ¾ 2 = 14 . Suppose the coin is tossed many times. Consider samples fx1 , x2 , x3 g of size n = 3, which are the numbers of heads obtained in each of three (independent) coin tosses. The associated sample mean x = X=

x1 + x2 + x3 has distribution 3

X1 + X2 + X3 3

= the average number of heads obtained in three coin tosses where X1 , X2 , X3 are identical copies of the distribution X. X is an estimator for the mean ¹ of X. For the sample f1, 0, 1g, corresponding to the coin toss results HTH, then x = estimate of ¹. Let pb be the sample proportion of heads obtained in n coin tosses. Then

X + X2 + :::: + Xn pb = 1 n

1+0+1 = 23 is an 3

pb is also a mean!

where X1 , X2 , ...., Xn are

identical copies of the distribution X.

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pb is an estimator for the population proportion p = 12 of heads obtained when an unbiased coin is tossed many, many times.

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83

STATISTICS AND PROBABILITY

For the sample f1, 1, 0, 0, 1, 0, 1g, corresponding to the coin toss results HHTTHTH, when the coin is tossed seven times, pb =

1+1+0+0+1+0+1 = 47 7

is an estimate of p.

Intuitively, a “good” estimator should have values which centre about the parameter it is approximating, and not be biased to values below or above the value of the parameter. The set of all estimates t, calculated from each possible sample using estimator T , is the sampling distribution of T . An estimator T for a population parameter µ is unbiased if the mean of the sampling distribution of T equals the parameter µ, so E(T ) = µ. Any estimate t from T is then called an unbiased estimate of µ. Otherwise, T is a biased estimator of µ, and any estimate t from T is then called a biased estimate of µ. Unbiased estimator T of µ

Biased estimator T of µ

T

µ

T

µ

or

T

µ

We have already proved in Section E that the sampling distribution of X has E(X) = ¹ and the sampling distribution of pb has E(pb) = p. Hence: ² A sample mean x is an unbiased estimate of the population mean ¹. ² A sample proportion pb is an unbiased estimate of the population proportion p. Suppose we have an unbiased estimator T for a population parameter µ, so E(T ) = µ, and therefore estimates found using T are centred about µ. The sampling distribution of T could be:

or T

µ

µ

T

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We see that the spread of the distribution of estimates could be small or large. It follows that the estimates found using T could vary but be close to µ, or could vary wildly from the value of µ.

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We therefore define: If T1 and T2 are two unbiased estimators for a parameter µ, then T1 is a more efficient estimator than T2 if Var(T1 ) < Var(T2 ). Example 43 Let X be a random variable with unknown mean ¹ and unknown variance ¾ 2 . Consider samples fx1 , x2 g of size 2 of independent values taken from X. Let T1 =

X1 + X2 3X1 + 5X2 and T2 = . 2 8

a Show that T1 is an unbiased estimator of ¹. b Show that T2 is an unbiased estimator of ¹. c Calculate estimates t1 and t2 for the sample f2:1, 3:5g. d Find Var(T1 ) and Var(T2 ). e Which of T1 and T2 is the more efficient estimator of ¹? Why? Since X1 and X2 each have distribution identical to the distribution of X, E(X1 ) = E(X2 ) = E(X) = ¹ and Var(X1 ) = Var(X2 ) = Var(X) = ¾ 2 . ³ ´ ³ ´ X + X2 3X1 5X2 a E(T1 ) = E 1 b E(T2 ) = E + 2

8

= 12 E(X1 + X2 )

= 12 (E(X1 ) + E(X2 ))

= 38 ¹ + 58 ¹ =¹ ) T2 is an unbiased estimator of ¹.

= 12 (¹ + ¹) =

8

= 38 E(X1 ) + 58 E(X2 )

2¹ 2

=¹ ) T1 is an unbiased estimator of ¹. c t1 =

2:1 + 3:5 5:6 = = 2:8 2 2

d Var(T1 ) = Var

¡1

2 X1

+ 12 X2

t2 =

¢

¡ ¢ Var(T2 ) = Var 38 X1 + 58 X2 ¡ ¢2 ¡ ¢2 = 38 Var(X1 ) + 58 Var(X2 )

= 14 Var(X1 ) + 14 Var(X2 )

= 14 ¾ 2 + 14 ¾ 2 =

2¾ 2 4

=

¾2 2

16 2 32 ¾


2. Example 44 Suppose X is a random variable with unknown mean ¹ and unknown variance ¾ 2 . Suppose f1:10, 2:5, 1:75, 3:45, 8:41, 6:75, 4:53g is a sample of independent values taken from X. a Calculate an unbiased estimate of ¹. i s62

b Calculate:

ii s72

c Write down E(S62 ) and E(S72 ) in terms of ¾ 2 .

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d Of the estimates of ¾ 2 found in b, which is the preferred estimate? Why?

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STATISTICS AND PROBABILITY

87

a The sample has size n = 7. ) x=

1:10 + 2:5 + 1:75 + 3:45 + 8:41 + 6:75 + 4:53 = 4:07 7

i sn2¡1 = s62

b

7 P

=

is an unbiased estimate of ¹.

ii sn2 = s72 = 67 s62

(xi ¡ x)2

¼ 6:19

i=1

6

¼ 7:21 c E(S62 ) = ¾ 2 and E(S72 ) = 67 ¾ 2 . d s62 is the preferred estimate of ¾ 2 since s62 is an unbiased estimate of ¾ 2 , whereas s72 is a biased estimate of ¾ 2 .

EXERCISE F 1 Suppose random samples of size 3 of independent values are taken from a population X with mean ¹ and variance ¾ 2 . a Show that T1 =

4X1 + 3X2 + 5X3 is an unbiased estimator of ¹. 12

b Show that T2 =

2X1 + X2 + 3X3 is an unbiased estimator of ¹. 6

c Which of T1 and T2 is the more efficient estimator? Why? a Consider a population with mean ¹ and variance ¾ 2 . Two independent random samples are taken with sizes 10 and 25, and the sample means x10 and x25 respectively are calculated. Which is the preferred estimate of ¹? Why?

2

b Hence explain why larger samples are better than smaller samples for estimating the population mean ¹. 3

a Suppose T1 and T2 are two independent unbiased estimators of a parameter µ. Show that T = aT1 + bT2 with a, b 2 R is an unbiased estimator of µ if and only if a + b = 1. b If T1 , T2 , ...., Tn are independent unbiased estimators of a parameter µ, under what condition(s) n P ai Ti , ai 2 R an unbiased estimator of µ? is T = i=1

4 A population with distribution X has mean ¹ and variance ¾ 2 . Random samples fx1 , x2 g of size 2 of independent values are taken from X. Let T = ¸X1 + (1 ¡ ¸) X2 , 0 6 ¸ 6 1, be an estimator of ¹. a Show that T is an unbiased estimator of ¹. b Use calculus to prove that the most efficient estimator of this form has ¸ = 12 . 5 Consider a population with unknown mean ¹ and unknown variance ¾ 2 . Two independent random samples fx1 , x2 , x3 , x4 g and fy1 , y2 , y3 , y4 , y5 , y6 , y7 g are taken. 4 7 P P 2 2

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is an unbiased estimate of ¾ 2 .

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a Show that t =

(yi ¡ y)

i=1

5

3

and sY2 =

100

(xi ¡ x)

i=1

5

Let sX2 =

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88

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b If the two samples had size n and m respectively, write down an unbiased estimate t of ¾ 2 in n m P P 2 2 terms of n, m,

sX2

=

(xi ¡ x)

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, and

n¡1

sY2

=

(yi ¡ y)

i=1

m¡1

6 Consider a normal distribution X with unknown mean ¹ and unknown variance ¾ 2 . Let X be the usual sample mean estimator of ¹, using random samples of n independent values.

. From Theorem 3, 2

2

Var(X ) = E(X ) ¡ fE(X)g

2

a Show that E(X ) > ¹2 . b What does the result in a imply?

7 Suppose X and Y are independent random variables with E(X) = ¹X , Var(X) = ¾ X2 , E(Y ) = ¹Y , and Var(Y ) = ¾ Y2 . n P A random sample of size n is taken from X, and the sample mean x and calculated.

sX2

=

(xi ¡ x)2

i=1 m P

sY2

n¡1

are

(yi ¡ y)2

= are Similarly, a sample of size m is taken from Y , and the sample mean y and m¡1 calculated. a Let U = X + Y . i Find E(U ) and Var(U ). ii Show that x + y is an unbiased estimate of E(U ). i=1

iii Show that sX2 + sY2 is an unbiased estimate of Var(U ). b Let i ii iii

U = aX + bY , where a, b 2 R + . Find E(U ) and Var(U ). Show that ax + by is an unbiased estimate of E(U ). Is asX2 + bsY2 an unbiased estimate of Var(U )? Explain your answer.

8 Let f2, 1:5, 6:78, 4:25, 8:61, 3:2g be a random sample of independent values taken from a population with unknown mean ¹ and unknown variance ¾ 2 . b Find an unbiased estimate of ¾ 2 .

a Find an unbiased estimate of ¹. c Calculate the sample variance sn2 .

d State the numeric relationship between your answers in b and c. Explain why we expect the value in b to be larger than the value in c. 9 Let X » U(0, b) be the continuous uniform random variable with probability density function (1 06x6b f(x) = b 0 otherwise. a Calculate explicitly E(X). b Let X be the sample mean estimator of E(X), calculated from random samples of size n of independent values of X.

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Show that 2X is an unbiased estimator of the parameter b, for all n 2 Z + .

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STATISTICS AND PROBABILITY

89

10 Suppose a random sample of size n of independent values is taken from a population with known mean ¹ and unknown variance ¾ 2 . n P 2 Show that

S¹2

=

(Xi ¡ ¹)

is an unbiased estimator of ¾ 2 .

i=1

n

11 Suppose random independent samples of independent values are taken from the same population with unknown mean ¹ and unknown variance ¾ 2 . n P 2 The variance of each sample is calculated using

sn2

=

(xi ¡ x)

i=1

n

.

Sample A of size 4 has sample variance sA2 = 3.

Sample B of size 9 has sample variance sB2 = 5.

Sample C of size 20 has sample variance sC2 = 2. a Let t =

4sA2 + 9sB2 + 20sC2 30

.

i Calculate the estimate t. ii Show that t is an unbiased estimate of ¾ 2 . b If r such samples with sizes n1 , n2 , ...., nr have sample variances s12 , s22 , ...., sr2 respectively, suggest a formula for an unbiased estimate t of ¾ 2 in terms of s12 , s22 , ...., sr2 . 12 Let X and Y be independent random variables with means ¹X and ¹Y respectively. Let x be the sample mean of a random sample of independent values taken from X. Let y be the sample mean of a random sample of independent values taken from Y . Show that the product of sample means x y is an unbiased estimate of ¹X ¹Y . ³ ´ p(1 ¡ p) 13 Consider a sample proportion pb » N p, . n

Let q = 1 ¡ p and qb = 1 ¡ pb, and let n be any constant, n 2 Z + . µ ¶ p bqb a Find E(pb) and E(qb). b Calculate E . n

c Hence explain why

p pq bqb is a biased estimate of the variance of pb. n n

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d Find an expression for an unbiased estimate of the variance

pq of pb. n

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90

STATISTICS AND PROBABILITY

G

CONFIDENCE INTERVALS FOR MEANS

It is often infeasible to calculate a population parameter. For example, consider calculating the mean weekly salary of all Spaniards aged 18 and over. In the previous section we looked at ways of estimating parameters, such as the population mean ¹, with approximate single values called point estimates. We want to know how confident we can be that our estimate is close to the true population parameter ¹. A confidence interval estimate of a parameter, in this case the population mean ¹, is an interval of values between two limits together with a percentage indicating our confidence that the true parameter ¹ lies in that interval. The Central Limit Theorem is used as a basis for finding the confidence intervals. For example, consider using sample means to estimate the population mean ¹. By the CLT, we can assume that approximately 95% of the sample means from samples of size n, lie within 2 standard errors of the population mean. E(X) = ¹, Var(X) =

¾2 , n

95.4%

¾ n

x1

¾X = p .

x2

x3

X

¹ 2¾X

The diagram shows the distribution of sample means, X.

2¾X

Consider the statement “We are 95% confident that the mean weekly salary of all adult Spaniards is between 637 euros and 691 euros”. The statement indicates that the population mean ¹ most likely lies in an interval between 637 euros and 691 euros. In particular, the probability that the interval contains the parameter ¹ is 0:95 .

THE 95% CONFIDENCE INTERVAL Consider the distribution X of all sample means x from samples of size n (large enough) taken from population X with population mean ¹ and population standard deviation ¾. X has distribution µ ¶ ¾2 ¾ X » N ¹, with mean ¹X = ¹ and standard deviation ¾ X = p . n

n

The corresponding standard normal random variable is Z =

X ¡¹

and Z » N(0, 1).

p¾

n

For a 95% confidence level we need to find a for which P(¡a 6 Z 6 a) = 0:95 .... (¤).

Area = 0.95

Using the symmetry of the graph of the normal distribution, the statement reduces to Area = 0.025 P(Z < ¡a) = 0:025 or P(Z < a) = 0:975. Using technology, we find that a ¼ 1:96 .

Area = 0.025 0

-a

Therefore, in (¤), P(¡1:96 6 Z 6 1:96) = 0:95 or P(¡1:96 6

X ¡¹ p¾

a

Z

6 1:96) = 0:95.

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91

STATISTICS AND PROBABILITY

Let x be any such sample mean. ¡1:96 6

x¡¹ p¾

6 1:96

n

¾ ¾ n n ¾ ¾ ) x ¡ 1:96 p 6 ¹ 6 x + 1:96 p n n

) ¡1:96 p 6 x ¡ ¹ 6 1:96 p

¾ n

¾ n

or equivalently ¹ ¡ 1:96 p 6 x 6 ¹ + 1:96 p

We therefore have the following equivalent results: ¾

¾

² 95% of all sample means x from samples of size n, lie between values ¹¡1:96 p and ¹+1:96 p n n inclusive. Area = 0.95 ) x lies in this area with probability 0.95.

¾ n

¹ ¡ 1:96 p

X

¾ n

¹ + 1:96 p

¹

² Given the sample mean x from one sample of size n, there is probability 0:95 that ¾ n

¾ n

¹ ¡ 1:96 p 6 x 6 ¹ + 1:96 p , and probability 0:05 that x lies outside this interval. Total area = 0.05 ) x lies in one of these areas with probability 0.05. ¾ n

¹ ¡ 1:96 p

X

¾ n

¹ + 1:96 p

¹

² Given the sample mean x from one sample of size n, there is probability 0:95 that the true population ¾ ¾ 6 ¹ 6 x + 1:96 p , and probability 0:05 that ¹ lies outside this n n

mean ¹ satisfies x ¡ 1:96 p interval.

¾

¾

The 95% confidence interval for ¹ using sample mean x is x ¡ 1:96 p 6 ¹ 6 x + 1:96 p , n n h i ¾ ¾ ¾ . also denoted x § 1:96 p or x ¡ 1:96 p , x + 1:96 p n

n

¾ n

n

¾ n

1:96 p

1:96 p

¾ n

x ¡ 1:96 p

¾ n

x + 1:96 p

x

lower limit

upper limit

We notice that: ² The exact centre or midpoint of the confidence interval is the value of x for the sample taken. ¾ n

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² The width of the 95% confidence interval for ¹ is 2 £ 1:96 p .

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STATISTICS AND PROBABILITY

² The use of 6 or < makes no difference in the calculation of the areas and therefore the probabilities for continuous random variables, but we require 6 in the definition of the confidence interval for later work on hypothesis testing.

OTHER CONFIDENCE INTERVALS FOR ¹ For a 90% confidence interval, P(Z < ¡a) = 0:05 or P(Z < a) = 0:95 .

Area = 0.90 Area = 0.05

Using technology, a ¼ 1:645. Since a is the

Area = 0.05 0

-a

Z

a

¾ n

coefficient of p in the confidence interval:

¾ ¾ 6 ¹ 6 x + 1:645 p n n

The 90% confidence interval for ¹ is x ¡ 1:645 p

Using this technique we obtain the following confidence intervals: Confidence level

a

Confidence interval

90%

1:645

x ¡ 1:645 p 6 ¹ 6 x + 1:645 p

95%

1:960

x ¡ 1:960 p 6 ¹ 6 x + 1:960 p

98%

2:326

x ¡ 2:326 p 6 ¹ 6 x + 2:326 p

99%

2:576

x ¡ 2:576 p 6 ¹ 6 x + 2:576 p

¾ n

¾ n

¾ n

¾ n

¾ n

¾ n

¾ n

¾ n

The confidence level is the amount of confidence we place in ¹ being within the calculated confidence interval.

We notice that: ² The sample mean x is the centre or midpoint of the confidence interval. ¾ where a is given in the table above. n

² The width of a confidence interval is 2 £ a £ p

² Increasing the sample size n produces confidence intervals of shorter width. For example, consider samples of different size but all with sample mean 10 and standard deviation 2. The 95% confidence interval is 10 ¡

1:960 £ 2 1:960 £ 2 6 ¹ 6 10 + p . p n n

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9:723 6 ¹ 6 10:277

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9:608 6 ¹ 6 10:392

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9:446 6 ¹ 6 10:554

0

50

5

9:123 6 ¹ 6 10:877

95

20

100

Confidence interval

50

n

75

25

0

5

For various values of n we have:

10.5

11

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STATISTICS AND PROBABILITY

INVESTIGATION 2

93

CONFIDENCE LEVELS AND INTERVALS

To obtain a greater understanding of confidence intervals and levels, click on the icon. The random sampler demonstration calculates confidence intervals at various levels of your choice (90%, 95%, 98%, or 99%) and counts the intervals which include the population mean.

DEMO

Example 45 A pharmaceutical company produces tablets with masses that are normally distributed with standard deviation 0:038 mg. A random sample of ten tablets was found to have mean mass 4:87 mg. Calculate a 95% confidence interval for the mean mass of these tablets, based on this sample.

¡ ¢ Even though n is relatively small, the fact that the mass X » N ¹, (0:038)2 is normally à µ ¶2 ! 0:038 p by the CLT. distributed ensures that X » N ¹, 10

Since x = 4:87, a 95% confidence interval for the mean mass ¹ of a tablet is 0:038

0:038

4:87 ¡ 1:96 £ p

6 ¹ 6 4:87 + 1:96 £ p

10

10

which is 4:846 6 ¹ 6 4:894 We are 95% confident that the population mean lies in the interval 4:85 6 ¹ 6 4:89.

Confidence intervals can be obtained directly from your graphics calculator. GRAPHICS CALCUL ATOR INSTRUCTIONS

CONFIDENCE INTERVALS FOR ¹ WHEN ¾ 2 IS UNKNOWN We usually do not know the population variance ¾ 2 , so instead we use sn2¡1 as an unbiased estimate of ¾ 2 as shown in Section F. If X is normally distributed then X is normally distributed, even for small sample size n. If X is not normally distributed, for sample size n large enough, the CLT says X is approximately normally distributed. X ¡¹

Now if ¾ 2 is known, Z =

p¾

» N(0, 1).

n

2

If ¾ is unknown, we use the random variable T =

sn2¡1 X ¡¹ sn¡1 p n

The Student’s t-distribution is named after William Gosset who wrote under the pseudonym “Student”.

instead, and for X normally distributed, has a t-distribution, sometimes called

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Student’s t-distribution.

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STATISTICS AND PROBABILITY

t-DISTRIBUTIONS standard normal curve N(0,¡1)

All t-distributions are symmetrical about the origin. They are like standardised normal bell-shaped curves, but with fatter tails. Each curve has a single parameter º (pronounced “new”) which is a positive integer. º is equal to the number of degrees of freedom of the distribution.

º=2 º = 10

0

For a given value of º, consider the t(º)-distribution. Denote by t® the value such that P(T > t® ) = ®, and equivalently P(T < ¡t® ) = ®.

Area = ®

Area = ® -t®

T ~ t(n)

t®

Since sn2¡1 has been calculated with n¡1 degrees of freedom, we find in general that

º = n ¡ 1.

For example, for a sample of size 8, º = 7, and we write T » t(7). The graphs illustrated are those of t(2), t(10) and Z » N(0, 1). As º = n ¡ 1 increases, the curve of the t(n ¡ 1) distribution approaches the standardised normal Z-curve. Consider taking random samples of size n from a distribution X with mean ¹ and unknown variance ¾ 2 . If X is normally distributed, it can be shown that T = n ¡ 1 degrees of freedom, and we write T » t(n ¡ 1).

X ¡¹ sn¡1 p n

follows a t-distribution with

In particular, when X is normally distributed with X » N(¹, ¾ 2 ), X is normally distributed for all values of n. ) T =

X ¡¹

has the t(n ¡ 1)-distribution with º = n ¡ 1 degrees of freedom.

sn¡1 p n

Suppose n, and therefore º = n ¡ 1, is fixed. Since P(¡t0:025 6

X ¡¹ sn¡1 p n

6 t0:025 ) = 0:95, the corresponding 95% confidence interval for ¹ is s

s

n¡1 n¡1 x¡ p t0:025 6 ¹ 6 x + p t0:025

n

n

sn¡1 with width 2 p t0:025 . n

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Other confidence intervals can be similarly defined.

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STATISTICS AND PROBABILITY

95

Example 46 The fat content, in grams, of 30 randomly selected pies at the local bakery was determined and recorded as: 15:1 14:8 13:7 15:6 15:1 16:1 16:6 17:4 16:1 13:9 17:5 15:7 16:2 16:6 15:1 12:9 17:4 16:5 13:2 14:0 17:2 17:3 16:1 16:5 16:7 16:8 17:2 17:6 17:3 14:7 Determine a 98% confidence interval for the average fat content of all pies made. Using technology, x ¼ 15:897 and sn¡1 ¼ 1:365. X is approximately normally distributed by CLT, since n = 30 is sufficiently large. X ¡¹

¾ is unknown and T =

sn¡1 p n

is T » t(29). GRAPHICS CALCUL ATOR INSTRUCTIONS

The 98% confidence interval for ¹ is s

s

n¡1 n¡1 x¡ p t0:01 6 ¹ 6 x + p t0:01

n

n 1:365 1:365 ) 15:897 ¡ p £ 2:462 6 ¹ 6 15:897 + p £ 2:462 30 30

) 15:283 6 ¹ 6 16:511 Alternatively, using technology, a 98% confidence interval for ¹ is 15:28 6 ¹ 6 16:51.

Example 47 A random sample of eight independent observations of a normal random variable gave P 2 x = 837:49 . Calculate: and

P

x = 72:8

a an unbiased estimate of the population mean b an unbiased estimate of the population variance, and hence an estimate of the population standard deviation c a 90% confidence interval for the population mean. P a x=

n

x

=

P

72:8 = 9:1 and so 9:1 is an unbiased estimate of ¹. 8

x2 837:49 ¡ x2 = ¡ 9:12 ¼ 21:876 n 8 n 8 An unbiased estimate of ¾ 2 is sn2¡1 = s 2 = £ 21:876 ¼ 25:00 n¡1 n 7

sn2

b

=

) an estimate of ¾ ¼ 5:00 c Using x = 9:1 and sn¡1 = 5:00, we obtain the 90% confidence interval for ¹. s

s

n¡1 n¡1 x¡ p t0:05 6 ¹ 6 x + p t0:05

n

n 5 5 ) 9:1 ¡ p £ 1:895 6 ¹ 6 9:1 + p £ 1:895 8 8

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) 5:750 6 ¹ 6 12:45 Alternatively, using technology, 5:75 6 ¹ 6 12:45 fusing the t-distributiong

STATISTICS PACKAGE

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96

STATISTICS AND PROBABILITY

DETERMINING HOW LARGE A SAMPLE SHOULD BE When designing an experiment in which we wish to estimate the population mean, the size of the sample is an important consideration. Finding the appropriate sample size is a problem that can be solved using the confidence interval. Example 48 Consider again Example 46 on the fat content of pies. Suppose the population standard deviation ¾ = 1:365 g. How large should a sample be if we wish to be 98% confident that the sample mean will differ from the population mean by less than 0:3 grams? We require ¡0:3 < ¹ ¡ x < 0:3 Now the 98% confidence interval for ¹ is ¾ ¾ ¹ 6 x + 2:326 p n n ¾ ¾ ) ¡2:326 p 6 ¹ ¡ x 6 2:326 p n n

x ¡ 2:326 p 6

The final answer for n needs to be rounded up here.

So, we need to find n such that ¾ n

2:326 p < 0:3 Consider

p 2:326¾ 2:326 £ 1:365 n= = ¼ 10:583 0:3

0:3

) n ¼ 112:01 So, a sample of size at least 113 should be taken to achieve the required accuracy.

EXERCISE G.1 1 The mean ¹ of a population is unknown, but its standard deviation is 10. In order to estimate ¹, a random sample of size n = 35 was selected. The mean of the sample was 28:9 . a Find a 95% confidence interval for ¹. b Find a 99% confidence interval for ¹. c In changing the confidence level from 95% to 99%, how does the width of the confidence interval change? 2 When performing a statistical analysis, we can choose the confidence level for a confidence interval. Why would statisticians not always choose to use confidence intervals of at least 99%? 3 A random sample of size n is selected from a population with known standard deviation 11. The sample mean is 81:6 . a Find a 95% confidence interval for ¹ if: i n = 36 ii n = 100. b In changing n from 36 to 100, how does the width of the confidence interval change? µ ¶ µ ¶ ¾ ¾ 4 The P % confidence interval for ¹ is x ¡ a p 6¹6x+a p . n

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5 A random sample of size n = 50 is selected from a population with standard deviation ¾. The sample mean is 38:7 . a Find a 95% confidence interval for the mean ¹ if: i ¾=6 ii ¾ = 15. b What effect does changing ¾ from 6 to 15 have on the width of the confidence interval? 6 Neville kept records of the time that he had to wait to receive telephone support for his accounting software. During a six month period he made 167 calls and the mean waiting time was 8:7 minutes. The shortest waiting time was 2:6 minutes and the longest was 15:1 minutes. a Estimate ¾ using ¾ ¼ range ¥ 6. Use the normal distribution to briefly explain why this estimate for ¾ is a reasonable one. b Find a 98% confidence interval for estimating the mean waiting time for all telephone customer calls for support. 7 A breakfast cereal manufacturer uses a machine to deliver the cereal into plastic packets. The quality controller randomly samples 75 packets and obtains a sample mean of 513:8 grams with sample standard deviation 14:9 grams. Construct a 99% confidence interval in which the true population mean should lie. 8 A sample of 42 patients from a drug rehabilitation program showed a mean length of stay on the program of 38:2 days with standard deviation 4:7 days. Estimate, with a 90% confidence interval, the average length of stay for all patients on the program. 9 A sample of 60 yabbies was taken from a dam. The sample mean weight of the yabbies was 84:6 grams, and the sample standard deviation was 16:8 grams. a For this yabbie population, find: i the 95% confidence interval for the population mean ii the 99% confidence interval for the population mean. b What sample size is needed to be 95% confident that the sample mean differs from the population mean by less than 5 g? 10 A random sample of ten independent observations of a normal random variable gave P 2 x = 1325:31 . Calculate: and

P

x = 112:5

a an unbiased estimate of the population mean b an unbiased estimate of the population variance, and hence an estimate of the population standard deviation c a 90% confidence interval for the population mean. 11 A porridge manufacturer knows that the population variance ¾ 2 of the weight of contents of each packet produced is 17:82 grams2 . How many packets must be sampled to be 98% confident that the sample mean differs from the population mean by less than 3 grams? 12 A sample of 48 patients from an alcohol rehabilitation program showed participation time on the program had a sample variance of 22:09 days2 . a Use the sample variance to estimate the population standard deviation ¾.

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b How many patients would have to be sampled to be 99% confident that the sample mean number of days on the program differs from the population mean by less than 1:8 days?

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98

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PAIRED DATA (THE CASE OF MATCHED PAIRS) Often we are interested in comparing sets of results for the same (or similarly matched) group(s) of individuals. For example, we might consider: ² race times for a class of students at the start and finish of the athletics season ² test results for two classes of the same size, of students of similar ability. In each case the data in the two samples obtained are matched in pairs. The two samples are not necessarily independent, for example a matched pair of race times for a particular individual at the start and finish of the athletics season. However, the individual scores in each sample must be independent for our analysis to be meaningful. We create a new single sample from the differences of the matched pairs and proceed with our usual methods for a single sample. For matched pairs, the population standard deviation ¾ will in general not be known. In such cases it is necessary to approximate ¾ by sn¡1 from the sample, and use the confidence interval for the t-distribution. Example 49 Prior to the 2004 Olympic Games an Institute of Sport took 20 elite athletes, and over a twelve month period monitored their training for the 100 m sprint. Below is the “best” time for each athlete in trials at the start and end of the year. The athletes have been recorded as the letters A to T, and times are in seconds. Athlete Start End

A 10:3 10:2

B 10:5 10:3

C 10:6 10:8

D 10:4 10:1

E 10:8 10:8

F 11:1 9:7

G 9:9 9:9

H 10:6 10:6

I 10:6 10:4

J 10:8 10:6

Athlete

K

L

M

N

O

P

Q

R

S

T

Start End

11:2 10:8

11:4 11:2

10:9 11:0

10:7 10:5

10:7 10:7

10:9 11:0

11:0 11:1

10:3 10:5

10:5 10:3

10:6 10:2

a Create i 95% ii 90% confidence intervals for the average time difference (start time ¡ end time) for all athletes in the relevant population. b The Institute of Sport claims their training program has improved sprint times. Do you agree? Explain your answer. a Let U = X1 ¡ X2 , where X1 represents the ‘start’ time and X2 represents the ‘end’ time. u

u

A 0:1

B 0:2

C ¡0:2

D 0:3

E 0

F 1:4

G 0

H 0

I 0:2

J 0:2

K

L

M

N

O

P

Q

R

S

T

0:4

0:2

¡0:1

0:2

0

¡0:1

¡0:1

¡0:2

0:2

0:4

Here n = 20, u = 0:155, and sn2¡1 = (0:344 085 36)2 .

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i The 95% CI for ¹ is ¡0:006 04 6 ¹ 6 0:316 . ii The 90% CI for ¹ is 0:021 96 6 ¹ 6 0:2880 . b An improvement in times corresponds to ¹ > 0. There is sufficient evidence at the 90% level that ¹ > 0, since ¹ = 0 lies outside the confidence interval, and ¹ is given to be positive. There is insufficient evidence at the 95% level to suggest that ¹ > 0, since the confidence interval contains negative values. Thus we would agree with the Institute of Sport’s claim at the 90% level, but not at the 95% level.

EXERCISE G.2 1 A group of 12 year old children were asked to throw a baseball as fast as they could. A radar was used to measure the speed of each throw. One year later, the same group was asked to repeat the experiment. The results are shown below, with the children labelled A to K, and the speeds given in km h¡1 .

Age

A

B

C

D

E

F

G

H

I

J

K

12 13

76 79

81 82

59 66

67 72

90 93

74 76

78 77

71 82

69 75

72 77

82 86

a Find confidence intervals for the average throwing speed difference for all children from age 12 i 95% confidence level ii 90% confidence level. to age 13, using a: b A sports commission report suggests that an average throwing speed difference of 5 km h¡1 is expected between these ages. Based on these experimental results, do you agree with the sports commission report? 2 Pairs of identical seedlings were grown with two types of compost, one with Type 1 compost and one with Type 2. The pairs were grown side by side in various garden plots. After a period of time, the height (in cm) of each seedling was measured. Pair Type 1

A 12:1

B 14:6

C 10:1

D 8:7

E 13:2

F 15:1

G 16:5

H 14:6

Type 2

12:3

15:2

9:9

9:5

13:4

14:9

17:0

14:8

a Determine unbiased estimates of the mean and variance of the difference d = (height Type 2) ¡ (height Type 1) for this paired data. b Calculate a confidence interval for ¹, the mean improved growth when using Type 2 compost i 95% ii 99%. instead of Type 1 compost, with confidence level: c The manufacturer of Type 2 compost guarantees it will improve the growth of seedlings more than Type 1 compost. Comment on this claim based on your calculations.

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d A confidence interval for ¹ is calculated as [¡0:032, 0:557]. Find, accurate to one decimal place, the % confidence level for this confidence interval.

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100

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H

SIGNIFICANCE AND HYPOTHESIS TESTING

Visitors to the West Coast of the South Island of New Zealand are often bitten by sandflies. According to its label, a new product claims to repel sandflies with “average protection time of more than 6 hours”. The best products currently available protect for six hours. The government department for tourism wishes to preserve the tourist trade, and therefore needs to provide the best possible advice to tourists. How can they test the manufacturer’s claim?

HYPOTHESES A statistical hypothesis is a statement about the value of a population parameter. The parameter could be a population mean ¹, or a proportion p. When a claim is made about a product, the claim can be tested statistically. The statistician begins by formulating a null hypothesis, H0 , that a parameter of the population takes a definite value, for example, that the population mean ¹ has value ¹0 . This statement is assumed to be true unless sufficient evidence is provided for it to be rejected. If the hypothesis is not rejected, we accept that the population mean is ¹0 , so the null hypothesis is a statement of no difference. The alternative hypothesis, H1 , is that there is a difference between ¹ and ¹0 . We will only accept this hypothesis if there is evidence to support it. The statistician then gathers a random sample from the population in order to test the null hypothesis. If the test shows that H0 should be rejected, then its alternative H1 is accepted.

ONE-TAILED AND TWO-TAILED ALTERNATIVE HYPOTHESES Given the null hypothesis H0 : ¹ = ¹0 , the alternative hypothesis could be: ² H1 : ¹ > ¹0 ² H1 : ¹ < ¹0 ² H1 : ¹ 6= ¹0

(one-tailed) (one-tailed) (two-tailed, as ¹ 6= ¹0 could mean ¹ > ¹0 or ¹ < ¹0 ).

For example, consider the case of the sandfly repellent: ² If the manufacturer of the new brand wants evidence that the new product is superior in protection time, the hypotheses would be: H0 : ¹ = 6 fthe new product has the same effectiveness as the old onesg H1 : ¹ > 6 fthe new product protects for longer than the old onesg.

The null hypothesis H0 always states ¹ equal to a specific value.

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² If the competitor wants evidence that the new product has inferior protection time, the hypotheses would be: H0 : ¹ = 6 fthe new product has the same effectiveness as the old onesg H1 : ¹ < 6 fthe new product protects for less time than the old onesg.

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² If a researcher studying all products on the market wants to show that the new product differs from the old ones, but is not concerned whether the protection time is more or less, the hypotheses would be: H0 : ¹ = 6 fthe new product has the same effectiveness as the old onesg H1 : ¹ 6= 6 fthe new product has different effectiveness from the old onesg.

ERROR TYPES There are two types of error in decision making: ² A Type I error is when we make the mistake of rejecting the null hypothesis H0 , when H0 is in fact true. ² A Type II error is when we make the mistake of accepting H0 when H0 is in fact not true. For example, if a coin is fair then the population proportion of heads it produces is p = 0:5 . ² A Type I error would be deciding a fair coin is biased because of the event of obtaining 10 heads in 10 tosses. Although improbable, it is still possible to obtain this result with a fair coin. ² A Type II error would be accepting that a biased coin is fair (when it is in fact biased) because of the event of obtaining 7 heads in 10 tosses. This event can occur with a fair coin with reasonable probability, but the coin used may in fact be biased towards heads.

EXERCISE H.1 1 Explain what is meant by: a a Type I error

b a Type II error

c the null hypothesis

d the alternative hypothesis.

a An experimenter wishes to test H0 : ¹ = 20 against H1 : ¹ > 20. i If the mean is actually 20 but the experimenter concludes that the mean exceeds 20, what type of error has been made? ii If the population mean is actually 21:8 but the experimenter concludes that the mean is 20, what type of error has been made?

2

b A researcher wishes to test H0 : ¹ = 40 against H1 : ¹ 6= 40. What type of error has been made if she concludes that: i the mean is 40 when it is in fact 38:1 ii the mean is not 40 when it actually is 40? 3 In many countries where juries are used in trials, “a person is presumed innocent until proven guilty”. In this case the null hypothesis would be H0 : the person on trial is innocent. a What would be the alternative hypothesis H1 ? b If an innocent person is judged guilty, what type of error has been made? c If a guilty person is judged as innocent, what type of error has been made?

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4 A researcher conducts experiments to determine the effectiveness of two anti-dandruff shampoos X and Y. He tests the hypotheses: H0 : X and Y have the same effectiveness H1 : X is more effective than Y. What decision would cause: a a Type I error b a Type II error?

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102

STATISTICS AND PROBABILITY

5 Current torch globes have a mean life of 80 hours. Globe Industries are considering mass production of a new globe they believe will last longer. a If Globe Industries wants to demonstrate that their new globe lasts longer, what set of hypotheses should they consider? b The new globe costs less to make, so Globe Industries will adopt it unless it has an inferior lifespan to the old type. What set of hypotheses would they now consider? 6 The top underwater speed of submarines produced at the dockyards is 26:3 knots. The engineers modify the design to reduce drag and believe that the maximum speed will now be considerably increased. What set of hypotheses should they consider to test whether or not the new design has produced a faster submarine?

HYPOTHESIS TESTING FOR THE MEAN WITH ONE SAMPLE WHERE ¾ IS KNOWN Consider a population X with unknown mean ¹ and known standard deviation ¾. Suppose we take a random sample of size n (large) of independent values of X, and calculate the sample mean x. We wish to test whether or not X has mean ¹ equal to a specific value ¹0 , based on this one sample mean x. The test can be performed as a two-tailed test, or as a one-tailed test. µ ¶ ¾2 , In either case we begin by assuming the null hypothesis H0 : ¹ = ¹0 . If this holds, then X » N ¹0 , and therefore Z =

X ¡ ¹0 p¾

n

has distribution N(0, 1), called the Null distribution.

n

For the given sample, we calculate z ¤ =

x ¡ ¹0 p¾

, called the test statistic.

n

Given the distributions for X and Z, we can calculate probabilities for where we expect a sample mean x to lie. If x lies in an extreme outer tail of the X-distribution, or equivalently z ¤ lies in an extreme outer tail of the Z-distribution, then either we have an extremely unlikely sample, or else the null hypothesis is incorrect. In this case we need to make a decision on whether or not to reject H0 and accept the alternative hypothesis H1 . EXAMPLE TWO-TAILED TEST Consider a two-tailed test with null hypothesis H0 : ¹ = ¹0 and alternative hypothesis H1 : ¹ 6= ¹0 . Assuming the population mean is indeed ¹0 , we know from the 95% confidence interval that for 95% of samples of size n, the sample mean x will satisfy ¾ n

Area = 0.95 ¾ n

¹ ¡ 1:96 p

¾ n

¾ n

¹ + 1:96 p

X

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STATISTICS AND PROBABILITY

¾ n

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¾ n

For the remaining 5% of samples, x will lie outside the interval [¹0 ¡ 1:96 p , ¹0 + 1:96 p ] in the outer tails of the distribution. We call these tails the critical region of the X-distribution, with boundary ¾ ¾ and ¹0 + 1:96 p as shown: n n

critical values ¹0 ¡ 1:96 p

critical region Area = 0.025

2:5% of sample means lie in the left section of the critical region. 2:5% of sample means lie in the right section of the critical region.

Area = 0.025 ¾ n

¹0 ¡ 1:96 p

X

¾ n

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¹0 + 1:96 p

If our particular sample mean x lies in the critical region, we can be 95% confident that ¹0 is not the population mean. There is 5% uncertainty, so we make the decision, at the 5% level of significance, to reject the null hypothesis H0 and accept the alternative hypothesis H1 . We note that there is probability 0:05 of making a Type I error. Although improbable, it is still possible that ¹0 is the true population mean, and we happened to select a random sample whose mean x had very low probability. EXAMPLE ONE-TAILED TESTS In the case of a one-sided alternative hypothesis, the critical region will be one tail only, and there is only one critical value. To calculate the critical value for a 5% level of significance, we need to find k such that P(Z > k) = 0:05 . We find k ¼ 1:645 . Hence, for a 5% level of significance, we have: One-tailed (right) test H0 : ¹ = ¹0

One-tailed (left) test H0 : ¹ = ¹0

H1 : ¹ > ¹0 Area = 0.05

¹0

H1 : ¹ < ¹0

Area = 0.05

X ¾ ¹0 + 1:645 p n

¾ ¹0 - 1:645 p n

X

¹0

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The level of significance determines the area of the critical region and therefore the critical values.

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TESTING PROCEDURE We have seen that the distribution of the test statistic z ¤ =

x ¡ ¹0 p¾

is Z » N(0, 1), called the null

n

distribution. CRITICAL REGIONS

For a level of significance ® which is the area of the critical region, we can calculate the critical value(s): One-tailed (left) test

One-tailed (right) test H0 : ¹ = ¹0

H0 : ¹ = ¹0

H1 : ¹ > ¹0

® = area of critical region

H1 : ¹ < ¹0

H0 : ¹ = ¹0

H1 : ¹ 6= ¹0

® = total area of critical region

® = area of critical region

Z

z®

Two-tailed test

Z

-z®

-z ®

Find z® by solving P(Z > z® ) = ®.

Find ¡z® by solving P(Z 6 ¡z® ) = ®.

For example, z0:01 ¼ 2:326

For example, ¡z0:01 ¼ ¡2:326

Z

z®

2

2

Find z ® by solving 2

P(Z > z ® ) = 2

® . 2

For example, z ® ¼ 2:576 2

and ¡z ® ¼ ¡2:576 2

In each case, x lies in the critical region of the X-distribution , z ¤ lies in the critical region of the null distribution.

p-VALUES Given our one sample mean x and corresponding test statistic z ¤ , we define the p-value to be the following probability: One-tailed (right) test

One-tailed (left) test

Two-tailed test

p = P(Z > z ¤ )

p = P(Z 6 z ¤ )

p = P(Z > jz ¤ j) + P(Z 6 ¡ jz ¤ j) = 2P(Z > jz ¤ j) fby symmetryg

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DECISION MAKING We decide to reject H0 in favour of H1 if any of the following (equivalent) properties hold: (1) The test-statistic z ¤ =

x ¡ ¹0 lies in the critical region of the null distribution Z » N(0, 1) ¾ p n

One-tailed (right) test H0 : ¹ = ¹0

One-tailed (left) test H0 : ¹ = ¹0

H1 : ¹ > ¹0

Two-tailed test

H1 : ¹ < ¹0

H0 : ¹ = ¹0

H1 : ¹ 6= ¹0

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area = ®

Z

z®

Z

-z®

z¤

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z¤

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z®

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2

z*

z* or

For example, for ® = 0:01, z ¤ > z0:01 ¼ 2:326

For example, for ® = 0:01, z ¤ < ¡z0:01 ¼ ¡2:326

For example, for ® = 0:01, z ¤ < ¡z0:005 ¼ ¡2:576 or z ¤ > z0:005 ¼ 2:576

(2) The p-value is strictly less than ® This is a comparison of the area of the tail(s) defined by z ¤ (or equivalently the area of the tail(s) defined by x), with the area of the critical region. One-tailed (right) test

One-tailed (left) test

Two-tailed test

Reject if p = P(Z > z ¤ ) < ®

Reject if p = P(Z 6 z ¤ ) < ®

Reject if p = 2P(Z > jz ¤ j) < ®

(3) x lies in the critical region of the X-distribution One-tailed (left) test

One-tailed (right) test

Two-tailed test total area = ®

area = ®

area = ®

X ¾ ¹0 + z® p n

X ¾ ¹0 - z® p n

x

x

x

X

x ¾ n

¹0 - z ® p 2

¾ n

¹0 + z ® p 2

For example, for ® = 0:01,

For example, for ® = 0:01,

For example, for ® = 0:01,

¾ x > ¹0 + 2:326 p n

¾ x < ¹0 ¡ 2:326 p n

¾ n ¾ or x > ¹0 + 2:576 p n

x < ¹0 ¡ 2:576 p

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Otherwise, we do not reject H0 as there is insufficient evidence to reject it. Note that although we accept H0 , we have not actually proved H0 . Rather, we simply have not found sufficient evidence against H0 .

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HYPOTHESIS TESTS AND CONFIDENCE INTERVALS For two-tailed tests, we can check the result of the hypothesis test using the appropriate confidence interval.

This rule is not true for one-tailed tests.

For example, For a two-tailed hypothesis test at the 5% level of significance, we accept H0 if and only if ¹0 lies within the 95% confidence interval for ¹.

USING A GRAPHICS CALCULATOR Click on the icon to obtain instructions for TI and Casio calculators. Be aware that your calculator may use different notation to that used in IB. For example:

GRAPHICS CALCUL ATOR INSTRUCTIONS

² with Casio calculators, sn¡1 is sx ² with TI calculators, sn¡1 is Sx .

REPORTING ON A HYPOTHESIS TEST There are effectively 7 steps in reporting on a hypothesis test: (1) Hypotheses:

State the null and alternative hypotheses. (Specify whether it is a one- or two-tailed test.)

(2) Null distribution: State the null distribution of the test statistic. (3) Test statistic:

Calculate the test statistic from the sample evidence.

(4) Decision rule:

State the decision rule based on the significance level ®.

(5) Evidence:

Find the p-value using your graphics calculator or find the critical values and the critical region.

(6) Decision:

Make your decision to reject or not reject H0 , based on the significance level.

(7) Conclusion:

Write a brief conclusion giving your decision some contextual meaning.

Example 50 The manager of a restaurant chain goes to a seafood wholesaler and inspects a large catch of over 50 000 prawns. It is known that the population standard deviation is 4:2 grams. She will buy the catch if the mean weight exceeds 55 grams per prawn. A random sample of 60 prawns is taken, and the mean weight is 56:2 grams. Is there sufficient evidence at a 5% level to reject the catch? Suppose the weight of prawns has distribution X with mean ¹ and ¾ = 4:2 g. µ ¶ (4:2)2 ) X » N ¹,

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H0 : ¹ = 55

(1) Hypotheses:

H1 : ¹ > 55 (one-tailed test)

(2) Null distribution: Z-distribution (¾ = 4:2 g is known) 56:2 ¡ 55

(3) Test statistic:

z¤ =

(4) Decision rule:

Reject H0 if the p-value is less than 0:05 .

or

(5) Evidence:

p-value = P(Z > 2:213) ¼ 0:0134

or

4:2 p

¼ 2:213

60

Reject H0 if z ¤ lies in the critical region of Z. area = 0.05

z0.05 = 1.645

Z z ¤ = 2.213

Since z ¤ lies in the critical region, we reject H0 .

(6) Decision:

Since the p-value is less than 0:05, we reject H0 .

(7) Conclusion:

Sufficient evidence exists at the 5% level to accept H1 , that the mean weight exceeds 55 grams. So, on this evidence, the manager should purchase the catch.

or

EXERCISE H.2 ¾ ¾ n n ¾ ¾ ¹ ¡ 1:96 p 6 x 6 ¹ + 1:96 p . n n

1 Show that if x ¡ 1:96 p 6 ¹ 6 x + 1:96 p then

a ® = 0:05

2 Find z® and z ® for: 2

b ® = 0:01

3 A population has known variance ¾ 2 = 15:79 . A sample of size 36 is taken and the sample mean x = 23:75 . We are required to test the hypothesis H0 : ¹ = 25 against H1 : ¹ < 25. a Find: i the test statistic

ii the null distribution

iii the p-value.

b What decision should be made at a 5% level using: i the test statistic ii the p-value? 4 For each of the following hypotheses, find the critical region for the test statistic for the standard i ® = 0:05 ii ® = 0:01 normal distribution with n > 30 and level of significance: a H0 : ¹ = 40 H1 : ¹ > 40

b H0 : ¹ = 50 H1 : ¹ < 50

c H0 : ¹ = 60 6 60 H1 : ¹ =

5 A statistician believes that a population which has a standard deviation of 12:9, has a mean ¹ that is greater than 80. To test this, he takes a random sample of 200 measurements, and the sample mean is 83:1 . He then performs a hypothesis test with significance level ® = 0:01 .

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6 Bags of salted cashew nuts display net contents 100 g. The manufacturer knows that the standard deviation of the population is 1:6 g. A customer claims that the bags have been lighter in recent purchases, so the factory quality control manager decides to investigate. He samples 40 bags and finds that their mean weight is 99:4 g. Perform a hypothesis test at the 5% level of significance, using critical regions, to determine whether the customer’s claim is valid. 7 An alpaca breeder wants to produce fleece which is extremely fine. In 2008, his herd had mean fineness 22:3 microns with standard deviation 2:89 microns. The standard deviation remains relatively constant over time. In 2012, a sample of 80 alpacas from the herd was randomly selected, and the mean fineness was 21:2 microns. a Perform a two-tailed hypothesis test at the 5% level of significance, using p-values to determine whether the herd fineness has changed. b Use a 95% confidence interval to check the result of your test. A machine packs sugar into 1 kg bags. It is known that the masses of the bags of sugar are normally distributed with a variance 2:25 g. A random sample of eight filled bags was taken and the masses of the bags measured to the nearest gram. Their masses in grams were: 1001, 998, 999, 1002, 1001, 1003, 1002, 1002. Perform a test at the 1% level, to determine whether the machine overfills the bags.

8

HYPOTHESIS TESTS WHERE THE POPULATION VARIANCE ¾ 2 IS UNKNOWN With ¾ 2 unknown, we can still define the null and alternative hypotheses as before. The distribution of µ ¶ ¾2 X is N ¹0 , . n

However, since ¾ 2 is unknown, we use the unbiased estimate sn2¡1 of ¾ 2 . We define the test statistic t¤ = distribution for this case.

x ¡ ¹0

which lies in the distribution t(n ¡ 1) called the null

sn¡1 p n

The critical region for t¤ and associated critical values in the t(n¡1)-distribution need to be calculated for the given value of n and given significance level ®. One-tailed (left) test

One-tailed (right) test

H0 : ¹ = ¹0

H1 : ¹ > ¹0 area = ®

T

-t ® 2

t®

T

2

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¡t® = invstudt (0:99, n ¡ 1) = ¡invstudt (0:01, n ¡ 1)

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For ® = 0:01, t ® = invstudt (0:005, n ¡ 1)

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For ® = 0:01, t® = invstudt (0:01, n ¡ 1)

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total area = ®

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t®

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H0 : ¹ = ¹0

H1 : ¹ < ¹0

area = ®

100

H0 : ¹ = ¹0

Two-tailed test

2

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We decide to reject H0 in favour of H1 if either of the following (equivalent) properties hold: (1) The test statistic t¤ =

x ¡ ¹0

lies in the critical region of the t(n ¡ 1)-distribution for the

sn ¡ 1 p n

given level of significance ®: One-tailed (left) test

One-tailed (right) test

T

t¤

t®

T

t¤ -t ®

t¤ > t®

Two-tailed test

t¤ -t ®

t¤ < ¡t®

T

t¤

t®

2

2

t¤ < ¡t ®

or t¤ > t ®

2

2

(2) The p-value is strictly less than ® This is a comparison of the area of the tail(s) defined by t¤ with the area of the critical region. The p-value in this case is defined as follows: One-tailed (right) test

One-tailed (left) test

area = p

Two-tailed test total area = p

area = p

T

t¤

T

t¤

p = P(T > t¤ )

p = P(T 6 t¤ )

¡ jt¤ j

jt¤ j

T

p = P(T > jt¤ j) + P(T 6 ¡ jt¤ j) = 2P(T > jt¤ j) fby symmetryg

Otherwise, we do not reject H0 as there is insufficient evidence to do so, and so we accept H0 . Example 51 In 2010, the average house price in a suburb was $235 000. In 2012, a random sample of 200 houses in the suburb was taken. The sample mean was x = $215 000, and an estimate of the standard deviation was sn¡1 = $30 000. Is there evidence at the 5% level that the average house price has changed? Suppose the price of a house has distribution X with mean ¹ and ¾ 2 unknown. µ ¶ ¾2 . ) X » N ¹, 200

H0 : ¹ = 235 000

(1) Hypotheses:

H1 : ¹ 6= 235 000 (two-tailed test)

215 000 ¡ 235 000

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¼ ¡9:428 with 199 degrees of freedom.

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30 p 000 200

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=

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sn¡1 p n

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Reject H0 if the p-value is less than 0:05 .

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(4) Decision rule:

0

t¤ =

5

95

x¡¹

(3) Test statistic:

100

50

75

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(2) Null distribution: t-distribution with º = 199, sn¡1 = 30 000 (¾ 2 is unknown)

Reject H0 if t¤ lies in the critical region of t.

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(5) Evidence:

p-value = P(T > 9:43) + P(T 6 ¡9:43) ¼ 1:11 £ 10¡17

or area = 0.025

area = 0.025

t0.025 ¼ 1.972

t¤ ¼ -9.43

T

-t0.025 ¼ -1.972

Since t¤ lies in the critical region, we reject H0 .

(6) Decision:

Since the p-value is less than 0:05, we reject H0 .

(7) Conclusion:

Sufficient evidence exists at the 5% level of significance, to suggest that ¹ 6= $235 000. We conclude that the average house price in 2012 was different to the average house price in 2010.

or

Example 52 Fabtread manufacture motorcycle tyres. Under normal test conditions, the average stopping time for motorcycles travelling at 60 km/h is 3:12 seconds. The production team have recently designed and manufactured a new tyre tread. They took 41 stopping time measurements under the usual test conditions, and found that the mean time was 3:03 seconds with sample standard deviation 0:27 seconds. Is there sufficient evidence, at a 1% level, to support the team’s belief that they have improved the stopping time? Suppose the motorcycle stopping time with the new tread has distribution X with mean ¹ and unknown ¾. µ ¶ ¾2 ) X » N ¹, . 41

H0 : ¹ = 3:12

(1) Hypotheses:

H1 : ¹ < 3:12 (one-tailed test)

(2) Null distribution: t-distribution with º = 40, sn2 = 0:272 (¾ 2 is unknown) sn2¡1 =

(3) Test statistic:

n 41 £ sn2 = £ 0:272 ¼ 0:074 72 n¡1 40

) sn¡1 ¼ 0:273 35 ) t¤ =

3:03 ¡ 3:12 0:273 p 35 41

¼ ¡2:108

(4) Decision rule:

Reject H0 if the p-value is less than 0:01 .

or

(5) Evidence:

p-value = P(T 6 ¡2:108) ¼ 0:020 67

or

Reject H0 if t¤ lies in the critical region of t. area = 0.01

T -t0.01 = -2.42

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Since t¤ is not in the critical region, we do not reject H0 .

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Since the p-value is not less than 0:01, we do not reject H0 .

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t¤ =

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There is insufficient evidence for us to reject H0 , so at the 1% level of significance, we retain H0 . We conclude that there is not an improvement in stopping time due to the new tread pattern.

(7) Conclusion:

EXERCISE H.3 1 Suppose º = 15. Find t® and t ® for:

a ® = 0:05

2

b ® = 0:01

2 A population has unknown variance ¾ 2 . A sample of size 24 is taken and the sample mean x = 17:14 with standard deviation 4:365 . We are required to test the hypothesis H0 : ¹ = 18:5 against H1 : ¹ 6= 18:5 . a Find: i the test statistic ii the null distribution iii the p-value. b What decision should be made at a 5% level using: i the test statistic ii the p-value? 3 A liquor store claimed that the mean price of a bottle of wine had fallen from what it was 12 months previously. Records show that 12 months ago the mean price was $13:45 for a 750 mL bottle. A random sample of prices of 389 different bottles of wine is now taken from the store. The mean price for the sample is $13:30, with sample standard deviation $0:25 . Is there sufficient evidence at a 2% level to reject the claim? In your answer state: a the null and alternative hypotheses

b the null distribution

c the test statistic e your conclusion.

d the p-value

4 A machine is used to fill bottles with 500 mL of water. Ten random measurements of the volumes put in different bottles give a mean of 499 mL with standard deviation 1:2 mL. Assuming that the volumes of water are normally distributed, test at the 1% level whether there is a significant difference from the expected value. 5 While peaches are being canned, 250 mg of preservative is supposed to be added by a dispensing device. To check the machine, the quality controller obtains 60 random samples of dispensed preservative. She finds that the mean preservative added was 242:6 mg with sample standard deviation 7:3 mg. a At a 5% level, is there sufficient evidence that the machine is not dispensing with mean 250 mg? Set out your solution in full, giving either a p-value or a critical value, and state your decision. b Use a confidence interval to verify your answer.

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6 Free range chickens are found to have mean meat protein content of 24:9 units per kg. 50 chickens are randomly chosen from a battery cage. These chickens are fed special meals which are supposed to increase their protein content. Following the feeding program, the sample had mean meat protein content 26:1 units/kg with standard deviation 6:38 units/kg. At a 5% level of significance, test the claim that the chickens on the feeding program have a higher meat protein content.

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7 The management of a golf club claimed that the mean income of its members was in excess of E95 000, so its members could afford to pay increased annual subscriptions. To show that this claim was invalid, the members sought the help of a statistician. The statistician was to examine the current tax records of a random sample of members fairly, and test the claim at a 0:02 significance level. The statistician found, from his random sample of 113 club members, that the average income was E96 318 with standard deviation E14 268. a Find an unbiased estimate of the population standard deviation. b State the null and alternative hypotheses when testing this claim. c State the null distribution. d Find the test statistic. e Find the p-value when testing the null hypothesis. f Find the critical region for rejection of the null hypothesis, and sketch it. g State whether or not there is sufficient evidence to reject management’s claim. h Would the statistician be committing a Type I or Type II error if his assertion was incorrect? i Find a 99% confidence interval for the mean income of members, and comment on your result. Explain why we check with a 99% confidence interval.

MATCHED PAIRS Example 53 Consider again the Institute of Sport problem in Example 49 on page 98. Prior to the 2004 Olympic Games an Institute of Sport took 20 elite athletes, and over a twelve month period monitored their training for the 100 m sprint. Below is the “best” time for each athlete in trials at the start and end of the year. The athletes have been recorded as the letters A to T, and times are in seconds. Athlete Start End

A 10:3 10:2

B 10:5 10:3

C 10:6 10:8

D 10:4 10:1

E 10:8 10:8

F 11:1 9:7

G 9:9 9:9

H 10:6 10:6

I 10:6 10:4

J 10:8 10:6

Athlete Start End

K 11:2 10:8

L 11:4 11:2

M 10:9 11:0

N 10:7 10:5

O 10:7 10:7

P 10:9 11:0

Q 11:0 11:1

R 10:3 10:5

S 10:5 10:3

T 10:6 10:2

Conduct a hypothesis test at the 5% level of significance, to determine whether the program has significantly improved the athletes’ performance. Let U = X1 ¡ X2 where X1 represents the time before and X2 represents the time after the program.

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u

K

L

M

N

O

P

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Here n = 20, u = 0:155, and sn¡1 = 0:344 085 36. (1) Null hypotheses:

H0 : ¹ = 0 (times have not improved) H1 : ¹ > 0 (one-tailed test as testing to see if times have improved)

(2) Null distribution: t-distribution (¾ 2 is unknown) t¤

(3) Test statistic:

=

u¡¹

area = 0.05

0:155 ¡ 0 = ¼ 2:014 56 0:0769

sn¡1 p n

T t¤ ¼ 2.015

t0.05 ¼ 1.73

(4) Decision rule:

Reject H0 if p-value is less than 0:05

(5) Evidence:

p-value ¼ P(T > 2:014 56) ¼ 0:029 16

(6) Decision:

Since the p-value is less than 0:05, we reject H0 . We observe that the test statistic t¤ ¼ 2:014 56 lies inside the critical region.

(7) Conclusion:

There is sufficient evidence at the 5% level, to conclude that the sprint times of the athletes have improved after the implementation of the program.

We have rejected the null hypothesis, yet the 95% confidence interval for ¹ does contain the value ¹ = 0. This is because we have a one-tailed test.

EXERCISE H.4 1 A mathematics coaching school claims to significantly increase students’ test results over a period of several coaching sessions. To test their claim a teacher tested 12 students prior to receiving coaching and recorded their results. The students were not given the answers or their results. At the conclusion of the coaching, the teacher gave the same test as before to check on the improvement. The paired results were: Student Before coaching

A 15

B 17

C 25

D 11

E 28

F 20

G 23

H 34

I 27

J 14

K 26

L 26

After coaching

20

16

25

18

28

19

26

37

31

13

27

20

Conduct a hypothesis test at a 5% level of significance, to see if the school’s claim was true. 2 Consider again the baseball problem on page 99: A group of 12 year old children were asked to throw a baseball as fast as they could. A radar was used to measure the speed of each throw. One year later, the same group was asked to repeat the experiment. The results are shown below, with the children labelled A to K, and the speeds given in km h¡1 . Age

A

B

C

D

E

F

G

H

I

J

K

12 13

76 79

81 82

59 66

67 72

90 93

74 76

78 77

71 82

69 75

72 77

82 86

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A sports commission report suggests that an average throwing speed difference of 5 km h¡1 is expected between these ages. Conduct a hypothesis test at a 5% level of significance to determine if the report’s claim is valid.

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3 Consider again the seedling problem on page 99: Pairs of identical seedlings were grown with two types of compost, one with Type 1 compost and one with Type 2. The pairs were grown side by side in various garden plots. After a period of time, the height (in cm) of each seedling was measured. Pair Type 1

A 12:1

B 14:6

C 10:1

D 8:7

E 13:2

F 15:1

G 16:5

H 14:6

Type 2

12:3

15:2

9:9

9:5

13:4

14:9

17:0

14:8

The manufacturer of Type 2 compost guarantees it will improve the growth of seedlings more than Type 1 compost. Based on these experimental results, conduct a hypothesis test at a 5% level, to determine whether the claim is valid.

THE PROBABILITY OF ERROR In our hypothesis tests, we have probability ® that for ¹ = ¹0 , our sample mean x could (validly) lie in the critical region of the X-distribution. The probability of making a Type I error = P(rejecting H0 when H0 is in fact true) = ®, the level of significance. Given that P(Type I error) = ®, we let P(Type II error) = ¯. In Example 52, the null hypothesis H0 is accepted since there is insufficient evidence to reject H0 . There is the chance here of making a Type II error, which is accepting H0 when H0 is in fact not true. If the significance level was ® = 0:05 and not 0:01, then H0 would have been rejected. Thus ¯ is dependent on ®. By reducing ® = P(Type I error), the probability ¯ = P(Type II error) is increased. We can only calculate ¯ = P(Type II error) if ¹, ¾, and n are all known.

In fact:

Suppose we conduct a hypothesis test with level of significance ® and null hypothesis H0 : ¹ = ¹0 , and suppose ¾ 2 is known. We accept H0 if the sample mean x does not lie in the critical region of the X distribution: One-tailed (right) test

One-tailed (left) test

X

¾ n

¹0 + z® p

x

Two-tailed test

X

¾ n

¹0 - z® p

x

X ¾ n

¹0 - z ® p 2

¾ n

¾ n

x 6 ¹0 + z® p

x > ¹0 ¡ z® p

x

¾ n

¹0 + z ® p 2

¾ n

¾ n

¹0 ¡z ® p 6 x 6 ¹0 +z ® p 2

2

But suppose H0 is false, and the true population mean is actually ¹ = ¹a , where ¹a 6= ¹0 .

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In this case, ¯ = P(Type II error) = P(accepting H0 : ¹ = ¹0 j ¹ = ¹a ) is calculated as follows:

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One-tailed (right) test test distribution

One-tailed (left) test

true distribution true distribution

¯ ¾ n

test distribution

X

¾ n

¹0 ¡ z ® 2

¾ n

¯ = P(X > ¹0 ¡ z® p )

¾2 ) n

where X » N(¹a ,

X

¹0 ¹a

¹0 - z® p

¯ = P(X 6 ¹0 + z® p )

true distribution

¯ ¾ n

¹0 + z® p

where X » N(¹a ,

test distribution

¯ ¹a ¹0

X

¹0 ¹a

Two-tailed test

¾ p n

¹0 + z ® 2

¾ p n

¾ n

2

2

¾ n

= 1 ¡ 2P(X > ¹0 + z ® p )

¾2 ) n

2

where X » N(¹a ,

¾2 ) n

The power of a test is defined to be 1 ¡ ¯, which equals the probability of (correctly) rejecting the null hypothesis H0 when H0 is false. Note that the closer ¹a is to ¹0 , that is the smaller j¹a ¡ ¹0 j is, the larger the value of ¯ = P(Type II error) = P(accept H0 j H0 is false), and hence the lower the power of the test.

¹0

X

¹a

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¯ critical value

¯

critical value

®

Example 54 Cans of chickpeas are labelled 400 g and it is known that the true weight of cans is normally distributed with a standard deviation of 10 g. A statistician wishes to conduct a test to see if the mean weight of a can is less than 400 g. He uses a sample of 12 cans, and a 2% level of significance. a Find the probability of: i a Type I error

ii a Type II error, given the true mean ¹ = 395 g.

b Find the power of the test, given the true mean ¹ = 395 g. i P(Type I error) = ® = 0:02 ii H0 : ¹ = 400 g H1 : ¹ < 400 g

a

10

H0 is retained if x > 400 ¡ 2:054 £ p

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¼ 394:0713

Area = 0.02

) ¯ = P(Type II error) = P(X > 394:0713)

10

-z0.02 = 400 ¡ 2:054 £ p

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Example 55 A sample of size 25 is taken from a normal population with unknown mean ¹ and known variance 36. To test the hypotheses: H0 : ¹ = 42, H1 : ¹ > 42 the decision rule is: accept H0 if the sample mean x 6 43:5 reject H0 if the sample mean x > 43:5 . a Find the probability of a Type I error for the given decision rule with critical value 43:5 . b Suppose the true value of the mean is 44:9 . i Find the probability of a Type II error. ii Suppose the critical value in the decision rule is changed. For what critical value is P(Type I error) = P(Type II error)? a X » N(¹,

36 25 )

) under the null hypothesis X » N(42, 1:22 ) ) P(Type I error) = P(rejecting H0 when H0 is true) = P(X > 43:5) ¼ 0:106

¾X = 1.2

42 43.5

i If ¹ = 44:9, then X » N(44:9, 1:22 ). We accept H0 if x 6 43:5 . ) P(Type II error) = P(accepting H0 when H0 is false) = P(X 6 43:5) ¼ 0:122

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¾X = 1.2

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42 + 44:9 = 43:45 2

equal areas

42

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44.9 43.45

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EXERCISE H.5 1 In a population of adult trout, it is known that the length X cm is normally distributed with known variance 6 cm2 , but the mean ¹ cm is unknown. It is proposed to test the hypotheses H0 : ¹ = 27 cm H1 : ¹ > 27 cm using the mean x of a sample of size 9. a Find the decision rule for the test, in terms of x, that corresponds to a significance level of i 5% ii 1%. b Suppose the true value of the mean is ¹ = 29:2 . Calculate P(Type II error) when P(Type I error) is i 0:05 ii 0:01 . 2 A sample of size 16 is taken from a normal population with unknown mean ¹ and known variance 64. The sample is used to test the hypotheses H0 : ¹ = 150, H1 : ¹ > 150. The decision rule for the test is: accept H0 if the sample mean x 6 155 reject H0 if x > 155. a Find P(Type I error). b Suppose the true mean is ¹ = 159. i Find the probability of a Type II error. ii Suppose a new critical value is used for the decision rule. Determine the critical value for x so that P(Type I error) = P(Type II error). 3 A sample of size 30 is taken from a normal population with unknown mean ¹ and known variance 7:5 . The sample is used to test the hypotheses H0 : ¹ = 37, H1 : ¹ < 37. a Determine the decision rule, in terms of the sample mean x, for the test so that P(Type I error) = 0:05 . b For the case where the true mean is ¹ = 36, find the power of the test. c Find the true value of ¹ given that P(Type I error) = 0:05 and P(Type II error) = 0:1 for this test. 4 The weight of a pumpkin from a very large crop is normally distributed with unknown mean ¹ and standard deviation 0:7 kg. Using a random sample of 15 pumpkins from the crop, a statistician conducts a hypothesis test at the 5% level of significance. Suppose that the true value of the mean is actually 6:4 kg. a For the test ¹ > 6 kg, calculate the probability of a Type II error. b For the test ¹ 6= 6 kg, calculate the power of the test. c For the test ¹ < 6 kg, calculate the power of the test. 5 The length of a beam produced in a manufacturing process is normally distributed with standard deviation 0:15 m. The manufacturer claims the mean length of such beams is 3:5 m. A random sample of 20 beams is taken and their mean length x is calculated. The value x is used to test the manufacturer’s claim. a State suitable hypotheses for a two-tailed test.

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b For a level of significance of 1%, define the critical region for x. c Calculate the probability of making a Type II error if the true mean length is 3:4 m.

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ERROR ANALYSIS WITH OTHER DISTRIBUTIONS Consider a hypothesis test where the null hypothesis H0 and alternative hypothesis H1 are defined and the decision rule for accepting or rejecting H0 is given either numerically or in abstract terms. We may be required to use any one of our known probability distributions to calculate: P(Type I error) = P(Reject H0 j H0 is true) = ® = level of significance for the test. P(Type II error) = P(Accept H0 j H1 is true) = 1 ¡ (power of the test). There are many possible questions of this form. We provide here only a few examples and exercises to demonstrate the general approach to such problems. The following results may be useful: Suppose X1 , X2 , X3 , ...., Xn are independent discrete random variables, and that Sn = X1 + X2 + X3 + :::: + Xn . ² If Xi are Bernoulli random variables Xi » B(1, p), then Sn » B(n, p). ² If Xi are Poisson random variables Xi » Po(¸i ), then Sn » Po(¸1 + ¸2 + ¸3 + :::: + ¸n ). ² If Xi are Geometric random variables Xi » Geo(p), then Sn » NB(n, p). ² If Xi are Negative binomal random variables Xi » NB(ri , p), then Sn » NB(r1 + r2 + r3 + :::: + rn , p). Example 56 To test whether a coin is fair, the following decision rule is adopted: Toss the coin 180 times. If the number of heads obtained X is between 75 and 105 inclusive, accept the hypothesis that the coin is fair, otherwise reject it. a Define the null hypothesis H0 and alternative hypothesis H1 . b

i Define a Type I error. ii Find the probability of making a Type I error. iii What is the level of significance for this test?

c Suppose the decision rule is changed so that P(Type I error) ¼ 0:05 . What is the new decision rule? d The coin is actually biased, and the probability of obtaining a head with each toss is 0:65 . Using the original decision rule, find P(Type II error). a H0 : The coin is fair, so p = probability of obtaining a head in one coin toss = 0:5 H1 : The coin is biased, so p 6= 0:5.

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i A Type I error is rejecting H0 when H0 is true. This means deciding the coin is biased when it is in fact fair. ii P(Type I error) = P(Reject H0 j H0 is true) = P(X 6 74 or X > 106 j p = 0:5) fwhere X » B(180, 0:5)g = 1 ¡ P(75 6 X 6 105) = 1 ¡ [P(X 6 105) ¡ P(X 6 74)] ¼ 0:0206

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iii The test is at about the 2% level of significance. c Assuming H0 is true, X » B(180, 0:5), and for the hypotheses in a we have a two-tailed test, where the total area of the critical region is 0:05 . Solving P(X 6 k) = 0:025 gives k = 77 For this discrete distribution, Solving P(X 6 k) = 0:975 gives k = 103 Check:

this is as close as we can get to 0:05 without exceeding it.

1 ¡ P(77 6 X 6 103) = 1 ¡ (P(X 6 103) ¡ P(X 6 76)) ¼ 0:0439

The new decision rule is: Toss the coin 180 times. If the number of heads obtained X satisfies 77 6 X 6 103, accept the null hypothesis that the coin is fair, otherwise reject it. d If p = 0:65, then X » B(180, 0:65). P(Type II error) = P(75 6 X 6 105 j X » B(180, 0:65)) ¼ 0:0374

EXERCISE H.6 1 A tetrahedral die has faces marked 1, 2, 3, and 4. To test whether the die is fair for rolling a 4, the following decision rule is adopted: Roll the die 300 times. If the number of 4s obtained, X, is between 62 and 88 inclusive, we accept the hypothesis that the die is fair for rolling a 4. Otherwise, we reject it. a Define the null hypothesis H0 and the alternative hypothesis H1 . b

i Define a Type I error for this example. ii Find the probability of making a Type I error. iii What is the level of significance for the test?

c Suppose the decision rule is changed so that P(Type I error) ¼ 0:02 . What is the new decision rule? d The die is actually biased with P(rolling a 4) = 0:32 . Use the original decision rule to find P(Type II error). 2 A machine fills each can of fizzy drink with volume Y cm3 , where Y is normally distributed with mean ¹ and standard deviation 2 cm3 . The mean ¹ is believed to be 330 cm3 . In order to check this value, a random sample of 16 cans is selected, and the sample mean y is calculated. The following hypotheses are set up: H0 : ¹ = 330 H1 : ¹ 6= 330. The critical region is defined as fy < 329g [ fy > 331g.

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a Find the significance level for this test. b If the true value of ¹ is found to be 328 cm3 , find the probability of a Type II error with this test.

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Example 57 A random variable X representing the number of successes in 270 trials can be modelled by a binomial distribution with parameters n = 270 and p, whose value is unknown. A significance test is performed, based on a sample value of x0 , to test the null hypothesis p = 0:6 against the alternative hypothesis p > 0:6 . The probability of making a Type I error is 0:05 . a Find the critical region for x0 . b Find the probability of making a Type II error in the case when p is actually 0:675 . a H0 : p = 0:6, H1 : p > 0:6 . Since n = 270 is very large, the binomial random variable X can be approximated by a normal random variable Xc » N(np, np(1 ¡ p)) where np = 270 £ 0:6 and np(1 ¡ p) area = 0.05 = 162 = 270 £ 0:6 £ 0:4 = 64:8 ) Xc » N(162, 64:8) The critical region is Xc > 175:2 But X is discrete, ) the critical region is X > 176.

162 175.2

X

b If p = 0:675, np = 270 £ 0:675 and np(1 ¡ p) = 182:25 = 270 £ 0:675 £ 0:325 ¼ 59:231 Now X » B(270, 0:675) can be approximated by Xc » N(182:25, 59:231), and from a the critical region is X > 175:2 ) P(Type II error) = P(H0 is accepted j H1 is true) = P(X 6 175:2 j p = 0:675) ¼ 0:180 or 18:0%. Example 58 A discrete random variable X has a Poisson distribution with unknown mean m. We wish to test the hypothesis H0 : m = 2 against H1 : m 6= 2. A random sample fx1 , ...., x12 g of 12 independent values is taken from X, with replacement. 12 P The decision rule is: accept H0 if 16 6 xi 6 33, otherwise reject it. i=1

a Define the critical region for S =

12 P

xi .

i=1

b

i Define a Type I error.

ii Calculate P(Type I error).

c The true value of m = 2:5 . i Define a Type II error.

ii Calculate P(Type II error).

a Since X is a discrete random variable with values 0, 1, 2, ...., so is S. ) the critical region is f0 6 S 6 15g [ fS > 34g.

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i A Type I error is rejecting H0 when H0 is in fact true. This means deciding m 6= 2 when in fact m = 2 and X » Po(2).

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ii Assuming H0 is true, X » Po(2) and ) S » Po(24). P(Type I error) = P(S is in the critical region j S » Po(24)) = P(0 6 S 6 15 or S > 34) = P(S 6 15) + 1 ¡ P(S 6 33) ¼ 0:034 40 + 1 ¡ 0:9686 ¼ 0:0658 i A Type II error is accepting H0 when H0 is in fact false. This means accepting m = 2 when in fact m = 2:5 . ii If X » Po(2:5), then S » Po(30). ) P(Type II error) = P(16 6 S 6 33)

c

= P(S 6 33) ¡ P(S 6 15) ¼ 0:743 3 Eri has a coin which has probability p of giving a head when tossed. She believes that the coin is fair. However, Eri’s friend, Mayuki, thinks that the coin is biased, and that p > 0:5 . To determine which of them is correct, Eri tosses the coin 12 times. Let X denote the number of heads obtained. a State appropriate null and alternative hypotheses. b Eri rejects the null hypothesis if X > 10. i What name is given to the region X > 10? ii What is meant by the significance level of a hypothesis test? Find its value for this test. c In fact, the coin is biased and p = 0:6 . Find the probability of a Type II error. d If Eri uses the decision rule from b, what type of error is Eri at risk of making? 4 To find the age of an archaeological specimen, a researcher measures the emission of radioactive particles. The number of particles X emitted in n minutes is said to have a Poisson distribution with parameter n¸, where the value of ¸ depends upon the age of the specimen. Two hypotheses concerning the age of one particular specimen are put forward: H0 : the specimen is 5000 years old, in which case ¸ = 2 H1 : the specimen is 10 000 years old, in which case ¸ = 5 . It is decided to count the number of radioactive particles X emitted in n minutes, and accept H0 if X 6 3, or reject H0 if X > 4. a Suppose n = 1. Find the probability of: i rejecting H0 given that H0 is true ii accepting H0 given that H1 is true.

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b In order that the researcher can publish his findings, the probability of accepting H0 given that H1 is true, must be less than 0:005. i Show that the minimum number of complete minutes for which counting should be recorded is three. ii Find the corresponding probability of rejecting H0 given that H0 is true in this case.

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Example 59 A discrete random variable X has a geometric distribution with unknown parameter p. We wish to test the hypothesis H0 : p = 0:3 against H1 : p 6= 0:3 . A random sample fx1 , x2 , ...., x10 g of independent values is taken from X, with replacement. ½ 10 ¾ ½ 10 ¾ P P xi 6 18 [ xi > 55 . The critical region is defined as i=1

a For which values of S =

10 P

i=1

xi will the null hypothesis be accepted?

i=1

b

i ii c The i

State the distribution of S, under the null hypothesis. Define a Type I error. iii Calculate the level of significance for this test. true value of p is p = 0:2 . Define a Type II error. ii Calculate the power of the test.

a X is a discrete random variable, and ) so is S. Hence the null hypothesis will be accepted for values of S such that 19 6 S 6 54. b

i If X » Geo(0:3) then S » NB(10, 0:3) has a negative binomial distribution. ii A Type I error is rejecting H0 when H0 is in fact true. This means deciding p 6= 0:3 when in fact p = 0:3 . iii The level of significance = P(Type I error) = P(S 6 18 or S > 55 j H0 is true) GRAPHICS CALCUL ATOR = P(S 6 18 or S > 55 j S » NB(10, 0:3)) INSTRUCTIONS = P(S 6 18) + (1 ¡ P(S 6 54)) µ ¶ 18 ¡¡ 54 ¡¡ ¢ ¢ ¢ ¢ P P i¡1 i¡1 (0:3)10 (0:7)i¡10 + 1 ¡ (0:3)10 (0:7)i¡10 = 9 9 i=10

¼ 0:040 26 ¼ 4% c

i=10

fusing technologyg

i A Type II error is accepting H0 when H0 is false. This means accepting p = 0:3 when in fact p = 0:2 . ii Power of the test = 1 ¡ P(Type II error) = 1 ¡ P(19 6 S 6 54 j p = 0:2) = 1 ¡ P(19 6 S 6 54 j S » NB(10, 0:2)) = 1 ¡ (P(S 6 54) ¡ P(S 6 18)) when S » NB(10, 0:2) ½ 54 ¾ 18 ¡ ¢ P ¡ i¡1 ¢ P i¡1 10 i¡10 10 i¡10 (0:2) (0:8) ¡ (0:2) (0:8) =1¡ 9 9 i=10

i=10

¼ 1 ¡ f0:660 39 ¡ 0:000 910 9g ¼ 0:341

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5 A box is known to contain either 20 white counters and 80 black counters, H0 , or 50 white counters and 50 black counters, H1 . In order to test hypothesis H0 against H1 , four counters are drawn at random from the box, without replacement. If all four counters are black H0 is accepted; otherwise it is rejected. a Find the probabilities of Type I and Type II errors for this test. b Determine whether the decision rule “If either three or four counters drawn are black, H0 is accepted; otherwise it is rejected” gives a test with more power. 6 A magician claims that he can roll a six with a fair die on average nine times out of ten. a Calculate the probability that he will roll five or more sixes in six rolls, assuming: i his claim is true ii he can roll a six, on average, only once in every six rolls. b To test the magician’s claim, he is invited to roll the die six times, his claim being accepted if he rolls at least four sixes. Find the probability that the test will: i accept the magician’s claim when hypothesis a ii is true ii reject the claim when it is justified, that is, when hypothesis a i is true. 7 A random variable X has a Poisson distribution with mean m, where m equals 3 or 4. To test the value of m the following hypotheses are defined: H0 : m = 3 H1 : m = 4. A random sample fx1 , x2 , ...., x9 g of independent values is taken from X, with replacement. 9 P xi 6 37, then H0 is accepted, otherwise H1 is accepted. If i=1

a Find the level of significance for this test.

b Calculate the power of the test.

8 A random variable X is known to have a geometric distribution with parameter p which is either 0:25 or 0:38 . To test the value of p, two hypotheses are defined: H0 : p = 0:25 H1 : p = 0:38 . A random sample fx1 , x2 , ...., x12 g of independent values is taken from X, with replacement. 12 P Let S = xi . If 30 6 S 6 71, then H0 is accepted, otherwise H1 is accepted. i=1

a Define the critical region in terms of S.

b Find P(Type I error).

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c Find P(Type II error).

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BIVARIATE STATISTICS

In this section we consider two variables X and Y which are counted or measured on the same individuals from a population. Since two variables are being considered, this is bivariate statistics. For example, for the given population, variables X and Y could be: Population

X

Y

A class of students

A student’s mark in Physics

A student’s mark in Mathematics

A collection of plants of the same species

The length of the stem of the plant

The distance of the plant from a water source

A class of students

A student’s mark in Drama

A student’s mark in Mathematics

We are interested in the following questions: ² Are X and Y dependent or independent? ² If X and Y are dependent, what is the nature of the relationship between X and Y ? ² Is the relationship between X and Y linear? If so, we can write Y = a + bX or X = c + dY for some constants a, b, c, d. ² If the relationship between X and Y is approximately linear, what is the strength of the linear dependence? ² Can we construct a reliable linear model Y = a + bX (or X = c + dY ) and use it to predict a value of Y given a value of X, or a value of X given a value of Y ?

CORRELATION AND CAUSATION Intuitively, correlation between two variables X and Y implies some sort of dependence. As one variable changes, the other variable also changes. However, correlation does not imply causation. Although two variables might increase or decrease in a related way, it is not necessary that a change in one variable causes a change in the other variable. It is possible there is something else influencing both X and Y , or it may simply be coincidence. For example, for a given population, it is observed over a period of time that: ² the number X of violent computer games sold per year increases ² the number Y of juvenile criminal convictions per year increases ² the number W of cupcakes sold per year increases. We may find a correlation between X and Y , and we may see a link whereby a change in X might cause a change in Y . However, proving causation requires much further analysis. In contrast, it would appear unreasonable to suggest that a change in X might cause a change in W .

SCATTER DIAGRAMS Consider two variables X and Y measured on the same population. Each of X and Y has a distribution of values. If there are n individuals being considered, we let xi and yi be the values of X and Y respectively measured on the ith individual.

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The n pairs f(x1 , y1 ), (x2 , y2 ), ...., (xn , yn )g form the joint probability distribution of X and Y . The graph of the points f(x1 , y1 ), (x2 , y2 ), ...., (xn , yn )g is called a scatter diagram (or scatter plot).

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The scatter diagram may be used to judge whether X and Y are independent, or whether there is a linear relationship between them. If all the points in the scatter diagram lie near or on a straight line, we say there is a linear correlation between X and Y . Y

Y

or

X

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If Y tends to increase as X increases, we have a positive linear correlation. For example: X = a student’s mark in Physics Y = a student’s mark in Mathematics.

Y

X

If Y tends to decrease as X increases, we have a negative linear correlation. For example: Y

X = the length of the stem of a plant Y = the distance of the plant from a water source.

X

If there is no relationship between X and Y (linear or otherwise) then there is no correlation between X and Y . We say that they are uncorrelated. X and Y are likely to be independent. For example: Y

X = a student’s mark in Drama Y = a student’s mark in Mathematics.

X

If all points (x, y) lie on a straight line, then we have a perfect linear correlation. positive

negative

Y

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THE (SAMPLE) PRODUCT MOMENT CORRELATION COEFFICIENT R AND ITS OBSERVED VALUE r Consider a scatter diagram of the joint probability distribution (X, Y ) which includes a random sample of n paired values f(x1 , y1 ), (x2 , y2 ), ...., (xn , yn )g. Let y = Let x =

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xi

P2(x2, y2)

be the mean of the x-values.

i=1

x2 - x

The point M(x, y) is called the mean point. The lines Y = y and X = x are called the mean lines. They divide the scatter diagram into four quadrants.

3rd quadrant

4th quadrant X

x

For a point P (x, y), consider the quantity (x ¡ x)(y ¡ y), equal to the product of the signed deviations of the point from each of the mean lines. ² If (x, y) lies in the 1st or 3rd quadrants, then (x ¡ x)(y ¡ y) > 0. ² If (x, y) lies in the 2nd or 4th quadrants, then (x ¡ x)(y ¡ y) < 0. For the n independent pairwise measured values (x1 , y1 ), (x2 , y2 ), ...., (xn , yn ), of X and Y on a random sample of n individuals from a population, the observed value of the (sample) product moment correlation coefficient R, is n P (xi ¡ x)(yi ¡ y) n n P P i=1 r=s where x = n1 xi and y = n1 yi . n n i=1 i=1 P 2 P 2 (xi ¡ x)

i=1

(yi ¡ y)

i=1

Since there are many possible samples in a population, we note that xi has distribution Xi identical to X, and yi has distribution Yi identical to Y , for i = 1, ...., n. For any such sample, we define the (sample) product moment correlation coefficient to be: n P (Xi ¡ X)(Yi ¡ Y ) n n P P where X = n1 Xi and Y = n1 Yi . R = s i=1 n n i=1 i=1 P 2 P 2 (Xi ¡ X)

i=1

µ

xi ¡ x ¾X

¶µ

yi ¡ y ¾Y

¶

i=1

=

1 n

n P

i=1

n

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STATISTICS AND PROBABILITY

127

We see that r is the average of the product of the standardised signed deviations of point (xi , yi ) from the mean lines. Therefore: ² r > 0 implies most points lie in the 1st and 3rd quadrants of the scatter diagram. This corresponds to a positive linear correlation. ² r < 0 implies most points lie in the 2nd and 4th quadrants of the scatter diagram. This corresponds to a negative linear correlation. ² r = 0 implies the points are spread symmetrically amongst all four quadrants. This corresponds to the variables being uncorrelated (in the linear sense). Example 60 For each data set: i Draw a scatter diagram including the mean point M(x; y) and mean lines Y = y and X = x. ii Discuss whether there is a linear correlation between X and Y . iii Calculate the standard deviations ¾ X and ¾ Y . iv Calculate r. a

X Y

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X Y

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X Y

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6 22

8 10 12 14 24 22 16 6

a

i x = 8, y = 20 ) M(x, y) = (8, 20) The mean lines are X = x = 8 and Y = y = 20. ii There is a perfect negative correlation between X and Y . iii ¾ X = 4, ¾ Y = 10 ¶µ ¶ 7 µ X xi ¡ x yi ¡ y 1 iv r = 7 =

3 5

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i x = 10, y = 17 ) M(x, y) = (10, 17) The mean lines are X = x = 10 and Y = y = 17. ii There is a perfect positive correlation between X and Y . iii ¾ X = 6, ¾ Y = 8 iv Using a similar calculation to a iv, r = 1.

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2 4 6 8 10 12 14 ¾X ¾Y i=1 ©¡ 2¡8 ¢ ¡ 35¡20 ¢ ¡ 4¡8 ¢ ¡ 30¡20 ¢ ¡ 6¡8 ¢ ¡ 25¡20 ¢ ¡ 8¡8 ¢ ¡ 20¡20 ¢ 1 + 4 + 4 + 4 7 4 10 10 10 10 ¡ 10¡8 ¢ ¡ 15¡20 ¢ ¡ 12¡8 ¢ ¡ 10¡20 ¢ ¡ 14¡8 ¢ ¡ 5¡20 ¢ª + 4 + + 10 4 10 4 10

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128

STATISTICS AND PROBABILITY

c

d

i x = 4, y = 3 ) M(x, y) = (4, 3) The mean lines are X = x = 4 and Y = y = 3. ii There is no linear correlation between X and Y . The variables are possibly independent. iii ¾ X ¼ 1:206, ¾ Y ¼ 1:348 iv r = 0

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Y 5 4 3 2 1

M

0

i x = 8, y = 16 ) M(x, y) = (8, 16) The mean lines are X = x = 8 and Y = y = 16. ii There is a quadratic relationship between the variables, so they are dependent. However, there is no linear correlation between them. iii ¾ X ¼ 4, ¾ Y ¼ 6:928 iv r = 0

1

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X

x=8

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X

From the above example we observe: ² perfect positive linear correlation has r = 1 ² perfect negative linear correlation has r = ¡1 ² if there is no linear correlation between X and Y , then r = 0. This example shows the extreme cases, and shortly we will prove that

Negative correlation

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moderate positive correlation

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strong negative correlation

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¡0:95 < r 6 ¡0:87

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strong positive correlation

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very strong negative correlation

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129

STATISTICS AND PROBABILITY

We note that even for the case of no linear correlation, where r = 0, X and Y may be independent or dependent. For example: Y

Y X, Y dependent

X, Y independent

X

X

EXERCISE I.1 1

a Draw the scatter diagram for the bivariate data set 2 8

x y

2 10

2 12

2 14

3 8

3 10

3 12

4 8

4 10

5 8

b Hence determine whether there is positive, negative, or no correlation. c Draw the lines x = x and y = y on the diagram. Do these lines and the four quadrants confirm your decision in b? n 1 P d Use the formula r = (zxi )(zyi ) to show that r = ¡0:5 . n i=1

n P

a Use r = s

2

n P

(xi ¡ x)(yi ¡ y)

to show that r = sµ

i=1 n P

(xi ¡ x)

2

i=1

n P

(yi ¡ y)

2

i=1

xi yi ¡ n x y

i=1 n P

i=1

¶µ

xi2 ¡ nx2

n P

i=1

¶. yi2 ¡ ny 2

b Hence show that r = ¡0:5 for the data set in question 1. 3 Consider the following set of bivariate data about the average daily food consumption of 12 obese adults. Weight (x kg)

96

87

84

75

98

88

93

81

89

64

68

78

Food consumption (y £ 100 calories per day)

35

34

35

29

41

35

36

30

34

26

28

31

a Draw a scatter diagram for the data set. b Describe the correlation between x and y. c Calculate the product moment correlation coefficient. 4 A selection of students were asked how many phone calls and text messages they had received the previous day. The results are shown below. Student Phone calls received Text messages received

A 4 6

B 7 9

C 1 2

D 0 2

E 3 5

F 2 8

G 2 4

H 4 7

a Draw a scatter diagram of the data.

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b Calculate r. c Describe the linear correlation between phone calls received and text messages received.

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130

STATISTICS AND PROBABILITY

5 A basketballer takes 20 shots from each of ten different positions marked on the court. The table below shows how far each position is from the goal, and how many shots were successful: Position Distance from goal (x m)

A 2

B 5

C 3:5

D 6:2

E 4:5

F 1:5

G 7

H 4:1

I 3

J 5:6

Successful shots (y)

17

6

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13

9

a Draw a scatter diagram of the data. b c d e

Do you think r will be positive or negative? Calculate the value of r. Describe the linear correlation between these variables. Copy and complete: As the distance from goal increases, the number of successful shots generally ......

f Is there a causal relationship between these variables? 1 1

x y

6 Consider the data set:

2 3

3 4

a Calculate r for the data set. b Let u = 2x + 1 and v = 3x ¡ 4. i List the data (u, v) in a table.

ii Calculate r for this data.

c Let u = ¡3x + 5 and v = 3x ¡ 4. i List the data (u, v) in a table.

ii Calculate r for this data.

d Let u = 2x + 1 and v = ¡3y ¡ 1. i List the data (u, v) in a table.

ii Calculate r for this data.

e Let u = ¡2x + 1 and v = ¡3y ¡ 1. i List the data (u, v) in a table.

ii Calculate r for this data.

f Compare your answers from a to e. What do your results suggest?

CORRELATION BETWEEN TWO RANDOM VARIABLES X AND Y AND THE (POPULATION) PRODUCT MOMENT CORRELATION COEFFICIENT ½ So far we have considered a measure r for the strength of correlation between variables X and Y using observed values (x1 , y1 ), (x2 , y2 ), ...., (xn , yn ). Such a set of values may well be from a population of n individuals. However, in practice it is likely to be only a sample of values from a larger population on which variables X and Y are measured. We now establish the theory to examine linear correlation between random variables X and Y for a general population. Consider the distribution of X and the distribution of Y for the whole population. Let E(X) = ¹X

denote the mean value of X

E(Y ) = ¹Y

denote the mean value of Y

Var(X) = ¾ X2

denote the variance of X

¾ Y2

denote the variance of Y .

Var(Y ) =

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Motivated by our previous work, we consider a scatter diagram for pairs (x, y) for X and Y measured on each individual in the population.

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131

STATISTICS AND PROBABILITY

Let M(¹X , ¹Y ) be the mean point and x = ¹X , y = ¹Y be the mean lines.

Y 2nd quadrant 1st quadrant

The mean lines divide the scatter diagram into four quadrants.

M

¹Y

3rd quadrant 4th quadrant X

¹X

For a pair (x, y), the quantity (x ¡ ¹X )(y ¡ ¹Y ) gives a signed measure of the deviations of x and y from their respective means ¹X and ¹Y . The average value of these deviations, E((X ¡ ¹X )(Y ¡ ¹Y )), provides a measure of the strength of linear dependence between X and Y . The covariance of X and Y is Cov(X, Y ) = E((X ¡ ¹X )(Y ¡ ¹Y )). We note that: ² The larger jCov(X, Y )j the greater the linear dependence of X and Y . ² If Cov(X, Y ) is large and positive, the majority of points (x, y) lie in the first and third quadrants of the associated scatter diagram. X and Y have a positive linear correlation.

Y

¹Y

M

X

¹X

² If Cov(X, Y ) is large and negative, the majority of points (x, y) lie in the second and fourth quadrants of the scatter diagram. X and Y have a negative linear correlation.

Y

¹Y

M

¹X

X

Theorem 14 Cov(X, Y ) = E(XY ) ¡ E(X)E(Y ) Proof: Cov(X, Y ) = E((X ¡ ¹X )(Y ¡ ¹Y )) = E(XY ¡ ¹Y X ¡ ¹X Y + ¹X ¹Y ) = E(XY ) ¡ ¹Y E(X) ¡ ¹X E(Y ) + ¹X ¹Y fby Theorem 7 and since E(¹X ¹Y ) = ¹X ¹Y by Theorem 1g = E(XY ) ¡ ¹Y ¹X ¡ ¹X ¹Y + ¹X ¹Y

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= E(XY ) ¡ E(X)E(Y )

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132

STATISTICS AND PROBABILITY

Theorem 15 If X, Y are independent random variables, then Cov(X, Y ) = 0. If X, Y are independent, then E(XY ) = E(X)E(Y ) fTheorem 7g

Proof:

) Cov(X, Y ) = E(XY ) ¡ E(X)E(Y ) = E(X)E(Y ) ¡ E(X)E(Y ) = 0. Note that converse of Theorem 15 is not true. If Cov(X, Y ) = 0, it is not necessarily true that X and Y are independent. We have observed that if Cov(X, Y ) is large and positive, then X and Y have a positive linear correlation. However, since the definition of covariance very much depends on the given scales and values of X and Y , it is difficult to quantify “large”. We therefore standardise the value by defining: The (population) product moment correlation coefficient ½ of random variables X and Y is ½= p

Cov(X, Y )

Var(X)Var(Y )

Note that ½ =

.

E((X ¡ ¹X )(Y ¡ ¹Y )) ¾X ¾Y

µµ

=E

X ¡ ¹X ¾X

¶µ

Y ¡ ¹Y ¾Y

¶¶

= E(ZX ZY ) where ZX = Theorem 16

X ¡ ¹X Y ¡ ¹Y , ZY = , and ¾ X , ¾ Y are the standard deviations of X and Y respectively. ¾X ¾Y

For X and Y two random variables with finite variances, ½ = p

Cov(X, Y )

Var(X)Var(Y )

Let ZX =

Proof:

X ¡ ¹X Y ¡ ¹Y , ZY = ¾X ¾Y

satisfies ¡1 6 ½ 6 1.

where Var(X) = ¾ X2 and Var(Y ) = ¾ Y2 .

) E(ZX ) = E(ZY ) = 0. Also E(ZX2 ) = Var(ZX ) + (E(ZX ))2 = 1 + 02 =1 and similarly E(ZY2 ) = 1. Let U = tZX + ZY . Since U 2 > 0, E(U 2 ) > 0 ) E((tZX + ZY )2 ) > 0 ) E(t2 ZX2 + 2tZX ZY + ZY2 ) > 0 ) t2 E(ZX2 ) + 2tE(ZX ZY ) + E(ZY2 ) > 0

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STATISTICS AND PROBABILITY

The discriminant 4½2 ¡ 4 of this quadratic therefore satisfies

133

4½2 ¡ 4 6 0 ) ½2 6 1

) ¡1 6 ½ 6 1 It follows that the observed value r of the sample product moment correlation coefficient satisfies ¡1 6 r 6 1. This is because, for X with distribution fx1 , x2 , ...., xn g and Y with distribution fy1 , y2 , ...., yn g, for the n data pairs (x1 , y1 ), (x2 , y2 ), ...., (xn , yn ) we have E(X) = x, E(Y ) = y, Var(X) = ¾ X2 , Var(Y ) = ¾ Y2 and therefore r = ½ for this data. Theorem 17 Let X and Y be random variables with finite variance. 1 If X and Y are independent then ½ = 0. 2 X and Y have a perfect linear correlation if and only if ½ = §1. Proof: 1 If X and Y are independent then Cov(X, Y ) = 0 fTheorem 15g ) ½= p

Cov(X, Y )

Var(X)Var(Y )

= 0.

Since X and Y are independent, there is no linear correlation between them. 2 ()) If Y = aX + b, a, b 2 R , a 6= 0, then ¹Y = E(Y ) = E(aX + b) = aE(X) + b = a¹X + b and Var(Y ) = a2 Var(X) Also Y ¡ ¹Y = aX + b ¡ ¹Y = aX + b ¡ (a¹X + b) = a(X ¡ ¹X ) ) Cov(X, Y ) = E((X ¡ ¹X )(Y ¡ ¹Y )) = E(a(X ¡ ¹X )2 ) = aE((X ¡ ¹X )2 ) = aVar(X) ) ½= p

Cov(X, Y )

Var(X)Var(Y )

=p

aVar(X)

a2 (Var(X))2

=

a = §1 jaj

(() Conversely, suppose ½ = 1 or ¡1. From the proof of Theorem 16 this means E(U 2 ) = t2 + 2t½ + 1 = (t + 1)2 if ½ = 1 or (t ¡ 1)2 if ½ = ¡1.

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Case E(U 2 ) = (t + 1)2 Let t = ¡1, so U = ZY ¡ ZX . ) E(U 2 ) = 0 and E(U ) = E(ZY ¡ ZX ) = E(ZY )¡ E(ZX ) = 0 ) Var(U ) = E(U 2 ) ¡ (E(U ))2 = 0 ¡ 02 = 0

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134

STATISTICS AND PROBABILITY

Thus U takes only one value, the mean 0, with probability 1. ) 0 = ZY ¡ ZX ) 0=

Y ¡ ¹Y X ¡ ¹X ¡ ¾Y ¾X

Rearranging, there exists a linear relationship between X and Y : Y =

¾Y ¹ X + (¹Y ¡ ¾ Y X ) ¾X ¾X

Case E(U 2 ) = (t ¡ 1)2 Let t = 1, so U = ZX + ZY . ) E(U 2 ) = 0 and E(U ) = E(ZX + ZY ) = E(ZX )+ E(ZY ) = 0 ) Var(U ) = E(U 2 ) ¡ (E(U ))2 = 0 ¡ 02 = 0 Thus U takes only one value, the mean 0, with probability 1. ) 0 = ZX + ZY ) 0=

X ¡ ¹X Y ¡ ¹Y + ¾X ¾Y

Rearranging, there exists a linear relationship between X and Y : Y =¡

¾Y ¹ X + (¹Y + ¾ Y X ) ¾X ¾X

The (population) product moment correlation coefficient ½ is hence a measure of the strength of the linear correlation between random variables X and Y . Values of ½ near 0 indicate a weak correlation and values of ½ near §1 indicate a strong correlation. Values ½ = §1 indicate a perfect linear correlation. In practice, we usually have only a sample from a population, and the observed value r of the (sample) product moment correlation coefficient is used as an estimate for ½.

EXERCISE I.2 1 Use the formula given in Theorem 14 for the covariance Cov(X, Y ) of two random variables X and Y , to show that: a Cov(X, X) = Var(X)

b Cov(X, X + Y ) = Cov(X, X) + Cov(X, Y )

c If X = c a constant, then Cov(X, Y ) = 0 for any random variable Y . 2 For X and Y two random variables, find Cov(X + Y , X ¡ Y ). 3 For X and Y two random variables, show that Var(X + Y ) = Var(X) + Var(Y ) + 2Cov(X, Y ). 4 Suppose X and Y are two random variables. Show that if Y = mX + c for constants m and c with m 6= 0, then ½ = §1. 5 Let X and Y be two random variables with correlation coefficient ½. Let U = a + bX and V = c + dY be two new random variables, for a, b, c, d constants. Find the product moment correlation coefficient of U and V in terms of ½.

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6 If X and Y are random variables with ¾ X2 = 1, ¾ Y2 = 9, and ½ = 19 , find the exact value of the correlation coefficient between X and X + Y . Hint: Use the result of question 3.

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STATISTICS AND PROBABILITY

THE TWO LINES OF REGRESSION Y

Consider the sample data (x1 , y1 ), (x2 , y2 ), ...., (xn , yn ). When we have a strong linear correlation between X and Y , it is useful to fit a straight line model to the data. If we are given the value of one variable, we can use the model to predict the value of the other.

(xn, yn) (x2, y2)

Conventionally, the “line of best fit” for data is obtained by Gauss’ method of least squares, as follows.

(x1, y1) X

THE REGRESSION LINE OF Y ON X For this regression line we rely more heavily on the measured values of x. We use this line to predict a value of y for a given value of x.

y = a + bx dn (xn, yn)

Y

Consider using a line of the form y = a + bx to model the data. The constants a and b are chosen so that the sum of the squared vertical distances of points from the line is a minimum. This means we minimise n P d12 + d22 + :::: + dn2 = (yi ¡ a ¡ bxi )2 .

(x2, y2) d2 d1 (x1, y1)

X

i=1

Suppose x =

n P

1 xi n i=1

and y =

n 1 P yi . n i=1

Calculus can be used to show that n P a = y ¡ bx and b =

n P

xi yi ¡ n x y

i=1 n P i=1

=

xi2 ¡ n(x)2

(xi ¡ x)(yi ¡ y)

i=1

n P

. (xi ¡ x)2

i=1

The regression line of Y on X has equation (y ¡ y) = b(x ¡ x) 2 n 3 P 6 i=1(xi ¡ x)(yi ¡ y) 7 7 (x ¡ x). which is (y ¡ y) = 6 n 4 P 5

We use this line of best fit to predict a value of y for a given value of x.

(xi ¡ x)2

i=1

The constant b is called the regression coefficient for this line. THE REGRESSION LINE OF X ON Y For this regression line we rely more heavily on the measured values of y. We use this line to predict a value of x for a given value of y. Consider using a line of the form x = c + dy to model the data. The constants c and d are chosen so that the sum of the squared horizontal distances of points from the line is a minimum. This means we minimise n P (xi ¡ c ¡ dyi )2 . h12 + h22 + :::: + hn2 =

Y

x = c + dy hn

(xn, yn)

(x2, y2) h2

h1

(x1, y1) X

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136

STATISTICS AND PROBABILITY n 1 P xi n i=1

Suppose x =

and y =

n 1 P yi . n i=1

Calculus can be used to show that n P c = x ¡ dy

and d =

xi yi ¡ n x y

i=1 n P i=1

yi2 ¡ n(y)2

n P

=

(xi ¡ x)(yi ¡ y)

i=1

n P

. (yi ¡ y)2

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i=1

The regression line of X on Y has equation (x ¡ x) = d(y ¡ y) 2 n 3 P 6 i=1(xi ¡ x)(yi ¡ y) 7 7 (y ¡ y). which is (x ¡ x) = 6 n 4 5 P 2 (yi ¡ y)

i=1

The constant d is called the regression coefficient for this line. We notice that the mean point M(x, y) lies on both regression lines. Thus either the two lines are identical or M(x, y) is the unique point of intersection of the two regression lines. Example 61 Consider the data set from Exercise I.1 question 3, about the average daily food consumption of 12 obese adults. Weight (x kg)

96

87

84

75

98

88

93

81

89

64

68

78

Food consumption (y £ 100 calories per day)

35

34

35

29

41

35

36

30

34

26

28

31

a Find the regression line of y on x and the regression line of x on y. b Find x and y, the mean of each data set. c Use the appropriate regression line to estimate: i the food consumption of an obese adult who weighs 92 kg ii the weight of an obese adult whose food consumption is 3250 calories per day. a

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The regression line of y on x is y = 0:364x + 2:454. The regression line of x on y is x = 2:398y + 4:68.

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STATISTICS AND PROBABILITY

b The point of intersection of the lines is M(83:4, 32:8). )

137

x ¼ 83:4, y ¼ 32:8

c Using the regression line of y on x with x = 92 kg, y ¼ 0:364 £ 92 + 2:454 ¼ 35:942 We expect the food consumption will be about 3594:2 calories per day. d Using the regression line of x on y with y = 32:5, x ¼ 2:398 £ 32:5 + 4:68 ¼ 82:6 kg We expect the weight will be about 82:6 kg. Theorem 18 For a random sample of n independent paired values (x1 , y1 ), (x2 , y2 ), ...., (xn , yn ): 1 The regression line of Y on X and the regression line of X on Y are the same line if and only if r = §1, in other words, if and only if X and Y have a perfect linear correlation. 2 If r = 0 then the regression line Y on X has equation y = y (a horizontal line) and the regression line X on Y has equation x = x (a vertical line). Proof: n P

By definition, r = s

(xi ¡ x)(yi ¡ y)

i=1 n P

(xi ¡ x)

2

i=1

Define: sxy = sx2 = sy2 = ) r=

sxy sx sy

n P

. (yi ¡ y)

2

i=1

n 1 P (xi ¡ x)(yi ¡ y) to be the sample covariance n i=1

n 1 P (xi ¡ x)2 to be the sample variance of the x-values n i=1

n 1 P (yi ¡ y)2 to be the sample variance of the y-values. n i=1

and the regression line of Y on X is (y ¡ y) = b(x ¡ x) sxy (x ¡ x) sx2 ³ ´ s x¡x = xy sx sx

=

1 We multiply by to obtain sy

µ

y¡y sy

¶ =r

³ ³

Similarly the regression line of X on Y is

x¡x sx x¡x sx

´ .... (1)

´

µ =r

y¡y sy

¶ .... (2)

1 Comparing (1) and (2) we see the two lines are identical if and only if r =

1 r

) r = §1

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2 If r = 0, then using equations (1) and (2) we find the regression line of Y on X is y = y, and the regression line of X on Y is x = x.

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138

STATISTICS AND PROBABILITY

Suppose b is the regression coefficient of the regression line Y on X and d is the regression coefficient of the regression line X on Y .

Corollary:

Then r is the geometric mean of b and d.

Using the same definitions as above, bd =

Proof:

r=

s2 sxy sxy = 2xy 2 = r2 2 2 sx sy sx sy

p bd

Theorem 18 (2) verifies the intuitive results that when X and Y are uncorrelated: ² no matter what the value of X, the expected value of Y is E(Y ) = y. ² no matter what the value of Y , the expected value of X is E(X) = x.

EXERCISE I.3 1 The results of ten people in a Mathematics and a Physics test were: Mathematics (x)

13

10

8

14

6

11

10

5

12

13

Physics (y)

21

15

14

20

12

16

9

10

17

12

a Determine the product moment correlation coefficient and the equations of the two regression lines. b Estimate the Mathematics score of someone who obtained a Physics score of 11. c Estimate the Physics score of someone who obtained a Mathematics score of 18. d Which of the estimates do you expect to be more accurate? Explain your answer. 2 The cholesterol level in the bloodstream (x) and the resting heart beat (y) of 10 people are: Cholesterol level (x)

5:32

5:54

5:45

5:06

6:13

5:00

4:90

6:00

6:70

4:75

Resting heart beat (y)

55

48

55

53

74

44

49

68

78

51

a Determine the product moment correlation coefficient and the equations of the two regression lines. b Estimate the cholesterol level of someone with a resting heart beat of 60. c Estimate the resting heart beat of someone with a cholesterol level of 5:8 . 3 Eight students swim 200 m breaststroke. Their times y in seconds, and arm lengths x in cm, are shown in the table below: Length of arm (x cm)

78

73

71

68

76

72

63

69

Breaststroke (y seconds)

123:1

123:7

127:3

132:0

120:8

125:0

140:9

129:0

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a Determine the product moment correlation coefficient and the equations of the two regression lines. b Estimate the time to swim 200 m breaststroke for someone with arm length 65 cm.

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STATISTICS AND PROBABILITY

139

4 Consider the bivariate data given below, the first line of which gives the weight x in kg of 10 men and 9 women and the second line of which gives the total weight y of body fat for each person. Men

Women

Weight (x kg)

89

88

66

59

93

73

82

77

100

67

Body fat (y kg)

12

14

9

10

22

13

13

11

19

12

Weight (x kg)

57

68

69

59

62

59

56

66

72

Body fat (y kg)

17

22

24

18

18

15

16

22

24

a Calculate the product moment correlation coefficient for X and Y for: i the men ii the women iii the 19 people in the data set.

You may wish to use a spreadsheet.

b Determine the percentage of body fat w for each person in the data set. c Calculate the product moment correlation coefficient for X and W for: i the men ii the women iii the 19 people in the data set.

THE BIVARIATE NORMAL DISTRIBUTION For single continuous random variables, the normal distribution is a classic distribution for which we have many theorems and results. In this section we generalise the normal distribution to two continuous random variables X and Y . X and Y have a bivariate normal distribution, or we say X and Y are jointly normally distributed, if they have the joint probability density function 8 ·³ ´2 ³ ´³ ´ ³ ´2 ¸ 9 > > x¡¹X x¡¹X y ¡¹Y y ¡¹Y > > ¡ ¡ 2½ + < = ¾X ¾X ¾Y ¾Y 1 exp f (x, y) = p 2(1 ¡ ½)2 > > 2¼¾ X ¾ Y 1 ¡ ½2 > > : ; for ¡1 < x < 1, ¡1 < y < 1, and where X is normal with mean ¹X and standard deviation ¾ X ,

expfxg = ex

Y is normal with mean ¹Y and standard deviation ¾ Y , and ½ =

Cov(X, Y ) is the (population) product moment ¾X ¾Y

correlation coefficient of X and Y . We have seen that if X and Y are any two independent random variables, then by Theorem 15, Cov(X, Y ) = 0 and therefore ½ = 0. For X and Y having a bivariate normal distribution, the converse of this result is also true. Theorem 19

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For X and Y with a bivariate normal distribution, X and Y are independent if and only if ½ = 0.

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Proof: By the remarks preceeding this theorem we need only prove the (() case. Suppose X and Y have a bivariate normal distribution with correlation coefficient ½ = 0. The joint probability density function of X and Y becomes ( "µ ¶2 µ ¶2 #) 1 x ¡ ¹X y ¡ ¹Y 1 f(x, y) = exp ¡ 2 + 2¼¾ X ¾ Y

Ã

1 exp p ¾ X 2¼

=

¾X

(

µ

¡ 12

x ¡ ¹X ¾X

¶2 )! Ã

¾Y

(

1 exp p ¾ Y 2¼

µ

¡ 12

y ¡ ¹Y ¾Y

¶2 )!

= f(x)f(y) where f (x) is the PDF of a normal random variable X with mean ¹X and standard deviation ¾ X and f(y) is the PDF of a normal random variable Y with mean ¹Y and standard deviation ¾ Y . Hence any probability calculated in the joint distribution will have form ¶ Z y2 µZ x2 P(x1 6 X 6 x2 , y1 6 Y 6 y2 ) = f (x, y) dx dy y1 y2

Z =

x1 x2

µZ

y1

µZ =

¶ f (x)f(y) dx dy

x1

x2

¶ µZ f(x) dx

x1

y2

¶ f (y) dy

y1

= P(x1 6 X 6 x2 ) £ P(y1 6 Y 6 y2 ) Hence X and Y are independent. SHAPE OF THE BIVARIATE NORMAL DISTRIBUTION The bivariate normal distribution takes the shape of a 3-dimensional bell-shaped surface. Consider the following cases which describe how the different parameters affect the position, shape, and orientation of the surface. Alongside each is a typical scatterplot for X and Y that might result from that distribution. Case 1: ¹X = ¹Y = 0 ¾X = ¾Y ½=0

) The peak is above the origin. ) The surface is axially symmetrical. ) X and Y are independent. Y f(x, y)

X

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STATISTICS AND PROBABILITY

141

Case 2: ¹X 6= 0, ¹Y 6= 0 ) The peak is at M(¹X , ¹Y ). ¾X = ¾Y ½=0

Y

f(x, y) mY mX (mX, mY)

X

y

x

Case 3: ¹X , ¹Y ¾ X 6= ¾ Y

) The surface has lines of symmetry parallel to the X and Y axes.

½=0

Y f(x, y) mY mX

(mX, mY)

x

X

y

Case 4: 3-D BELL-SHAPED SURFACE

¹X , ¹Y ¾ X 6= ¾ Y ½ 6= 0 ) X and Y are now dependent. The surface has lines of symmetry not parallel to the X and Y axes. Y f(x, y) mY mX

(mX, mY)

x

X

y

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The result in Case 4 is a probability distribution which favours the points on the scatterplot for X and Y showing linear correlation. For a sufficiently large sample of points, the regression line will approximate one of the lines of symmetry of the bell-shaped probability distribution surface.

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142

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HYPOTHESIS TESTING FOR DEPENDENCE OF X AND Y We have seen that to analyse the strength of linear correlation between two random variables X and Y we use observed values (x1 , y1 ), (x2 , y2 ), ...., (xn , yn ), and either a scatter diagram or the value of r. In the case of X and Y having a bivariate normal distribution, a single such value of r can be used in a hypothesis test to determine (at the given level of significance) whether or not ½ = 0 for X and Y , and therefore whether or not X and Y are linearly correlated. By Theorem 19, this will also tell us whether or not X and Y are independent. We require the following result: If X and Y have a bivariate normal distribution such that ½ = 0, then the sampling distribution r n¡2 R has the Student’s t-distribution with (n ¡ 2) degrees of freedom. 2 1¡R

Hence for any random sample of n independent paired data values (x1 , y1 ), (x2 , y2 ), ...., (xn , yn ) from the jointly normal distribution (X, Y ) with ½ = 0, the observed value r of the sample product r n¡2 moment correlation coefficient is such that r lies in the t(n ¡ 2)-distribution. 2 1¡r

HYPOTHESIS TEST WITH H0 : ½ = 0 AND H1 : ½ 6= 0 Given a random sample of n independent paired data values (x1 , y1 ), (x2 , y2 ), ...., (xn , yn ) from a bivariate normal distribution (X, Y ): (1) Calculate r, the observed value of the sample product moment correlation coefficient. r n¡2 ¤ . (2) Calculate the test statistic t = r 2 1¡r

(3) Either For level of significance ®, calculate the critical values §t ® for the t(n ¡ 2)-distribution 2 which define the two-tailed critical region.

Area = ®_w

Area = ®_w -t ®

Calculate the p-value

or

T

t®

2

2

P(T 6 ¡jt¤ j) + P(T > jt¤ j) = 2P(T > jt¤ j) p

p

Area = _w

Area = _w

T

jt¤ j

¡ jt¤ j

(4) We do not reject H0 if ¡t ® 6 t¤ 6 t ® , or if the p-value is not less than ®. 2

2

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We reject H0 in favour of H1 if t¤ lies in the critical region t¤ < ¡t ® or t¤ > t ® , or if the 2 2 p-value is less than ®.

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STATISTICS AND PROBABILITY

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If the result of the hypothesis test is to reject H0 : ½ = 0 in favour of H1 : ½ 6= 0, we conclude that the bivariate data is correlated. The strength of this correlation is measured by the value of r. In this case the regression line of Y on X and the regression line of X on Y can be calculated, and predictions for values of X and Y can be made. Example 62 Consider the following data set sampled from a bivariate normal distribution. 4 3

x y

3 4

4 ¡3

3 ¡4

¡4 3

¡3 4

¡4 ¡3

¡3 ¡4

Conduct hypothesis tests at the 5% and 1% levels of significance, to determine whether or not the two sets of data are correlated. We test at the 5% and 1% levels: H0 : ½ = 0 against H1 : ½ 6= 0. n = 8, so there are 6 degrees of freedom. (1) r is calculated and is found to be zero. r r n¡2 8¡2 (2) t¤ = r = (0) = 0. 2 2 1¡r

1 ¡ (0)

t¤

(3) Either Since

= 0, ¡t ® 6 t¤ 6 t ® no matter what the level of significance is. 2

2

The p-value equals 1.

or

(4) We retain H0 at both 5% and 1% levels, and conclude X and Y are uncorrelated.

Example 63 The following data set is sampled from a bivariate normal distribution: 2 8

x y

2 10

2 12

2 14

3 8

3 10

3 12

4 8

4 10

5 8

Conduct a hypothesis test at the 5% level of significance to determine whether or not the two sets of data are correlated. H1 : ½ 6= 0 H0 : ½ = 0 (1) r = ¡0:5 r r n¡2 ¤ = ¡0:5 (2) t = r 2 1¡r

GRAPHICS CALCUL ATOR INSTRUCTIONS

10 ¡ 2 ¼ ¡1:633 1 ¡ (¡0:5)2

(3) n = 10 ) there are 8 degrees of freedom Either For ® = 0:05, t ® = t0:025 ¼ 2:306 2

Area = 0.025

) t¤ does not lie in the critical region.

Area = 0.025 T

-t ® = -2.306

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t¤ ¼ -1.633

t ® = 2.306 2

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144

STATISTICS AND PROBABILITY

p-value = 2P(T > jt¤ j) ¼ 2P(T > 1:633) ¼ 0:1411 ) p-value > ® (4) Using either the critical region test or the p-value, we do not reject H0 at the 5% level of significance. We therefore conclude that ½ = 0, which means the sets of data are uncorrelated, and X and Y are independent variables. or

EXERCISE I.4 1 Perform a hypothesis test on the following bivariate data to determine whether or not the variables are linearly correlated. Mathematics (x)

13

10

8

14

6

11

10

5

12

13

Physics (y)

21

15

14

20

12

16

9

10

17

12

Consider both 5% and 1% levels of significance, and use both the critical region test and the p-value. 2 The cholesterol level in the bloodstream (x) and the resting heart beat (y) of 10 people are: Cholesterol level (x)

5:32

5:54

5:45

5:06

6:13

5:00

4:90

6:00

6:70

4:75

Resting heart beat (y)

55

48

55

53

74

44

49

68

78

51

Perform a hypothesis test to determine whether or not the variables are linearly correlated. Consider both 5% and 1% levels of significance, and use both the critical region test and the p-value. 3 The bivariate data below is sampled from a bivariate normal distribution. X is the number of Sudoku puzzles solved in a three hour period, and Y is the number of logic puzzles solved in a three hour period. 5 3

x y

8 11

12 9

15 6

15 15

17 13

20 25

21 15

25 13

27 20

Carry out a hypothesis test to determine whether or not the variables are linearly correlated. Consider both 1% and 5% levels of significance, and use only the p-value. 4 Consider the following data sampled from a bivariate normal distribution. Weight of Mother (x kg)

49

46

48

45

46

42

43

40

52

55

68

Birth weight of child (y kg)

3:8

3:1

2:5

3:0

3:2

2:8

3:1

2:9

3:4

2:9

3:9

Carry out a critical region hypothesis test to determine whether or not the variables X and Y are independent, using: a a 5% level of significance

b a 1% level of significance.

5 A random sample of n = 16 paired values from a bivariate normal distribution (X, Y ) has correlation coefficient of r = ¡0:5 . Does this indicate the variables X and Y are correlated: a at the 5% level of significance b at the 1% level of significance?

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6 A random sample of paired values from a bivariate normal distribution has perfect correlation, so r = 1. Explain why this data cannot be used in a hypothesis test for the correlation of X and Y . Do you think a hypothesis test would be needed in this case?

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Example 64 A random sample with correlation coefficient r = 0:5 is taken from a bivariate normal population (X, Y ). What is the minimum sample size n required to conclude that X and Y are linearly correlated at the 1% level of significance? We test H0 : ½ = 0 against H1 : ½ 6= 0 with r = 0:5 r p n¡2 ¤ ) the test statistic is t = 0:5 ¼ 0:577 n ¡ 2 . 2 1 ¡ (0:5)

For infinitely many degrees of freedom the t-distribution approximates the normal distribution Z » N(0, 1) where t ® ¼ z ® = z0:005 ¼ 2:576. 2

2

Thus t¤ > t ® 2

p if 0:577 n ¡ 2 > 2:576 ) n > 21:9 ) we require n > 22

Try n = 22, º = 20, t0:005 ¼ 2:845, t¤ ¼ 2:580 ) t¤ < t0:005

) not significant.

n = 23, º = 21, t0:005 ¼ 2:831, t¤ ¼ 2:644 ) t¤ < t0:005

) not significant.

n = 24, º = 22, t0:005 ¼ 2:818,

t¤

¼ 2:706 )

t¤

< t0:005

) not significant.

n = 25, º = 23, t0:005 ¼ 2:807, t¤ ¼ 2:767 ) t¤ < t0:005

) not significant.

n = 26, º = 24, t0:005 ¼ 2:797, t¤ ¼ 2:827 ) t¤ > t0:005 ) for n > 26 we would reject H0 : ½ = 0 in favour of H1 : ½ 6= 0. ) a sample size of at least 26 is required to conclude, at the 1% level of significance, that X and Y are correlated. 7 Consider a random sample with correlation coefficient r = 0:6 taken from a bivariate normal distribution (X, Y ). What is the minimum sample size n necessary to decide that X and Y are linearly correlated at the 5% level of significance?

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8 Consider a random sample of 20 data pairs from a bivariate normal distribution (X, Y ). What is the least value of jrj for this sample for which we would conclude, at the 5% level of significance, that X and Y are correlated?

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REVIEW SET A 1 X1 , X2 , and X3 are random variables, each with mean ¹ and standard deviation ¾. a Find the mean and standard deviation of: ii 2X1 ¡ 3X2 + X3 i X1 + 2X2 + 3X3 ¡ ¢ 2 b Find E [X1 ¡ X2 ] given that X1 and X2 are independent. 2 A student is waiting at a bus stop. He knows that 35% of all the buses going past can take him to school. The other buses go elsewhere. a Suppose the student will catch the first bus that will take him to school. i Find the probability that it will take at most 4 buses for there to be a correct one. ii Find the average number of buses it will take for there to be a correct one. b Suppose the student decides that he will catch the third bus that would take him to school, because he thinks his friend will be on that bus. i Find the probability that he will catch the 7th bus that goes past. ii Find the average number of buses it will take for the correct bus to arrive. iii Find the probability that it will take no more than 5 buses for the correct bus to arrive. 3 The probability distribution for the random variable X is given in the table alongside. Find the: a value of c c probability that X is greater than the mean

X=x

¡3

¡1

1

3

5

P(X = x)

c

c

c

c

c

b mean of X d variance of X.

4 In the Japanese J-League, it it known that 75% of all the footballers prefer to kick with their right leg. a In a random sample of 20 footballers from the J-League, find the probability that: i exactly 14 players prefer to kick with their right leg ii no more than five prefer to kick with their left leg. b In a random sample of 1050 players from the J-League, find the probability that: i exactly 70% of players prefer to kick with their right leg ii no more than 25% prefer to kick with their left leg. Hint: For b, use a suitable approximation for the random variable X = the number of footballers who prefer to kick with their right leg. 5 Suppose X » B(n, p) where x = 0, 1, 2, 3, 4, ...., n. a Show that X has Probability Generating Function G(t) = (1 ¡ p + pt)n . i E(X) = np

b Hence prove that:

ii Var(X) = np(1 ¡ p).

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6 To estimate the mean number of hours that employees are away from work in a year due to sickness, a sample of 375 people is surveyed. It is found that for last year, the standard deviation for the number of hours lost was 67. Suppose we use this to approximate the standard deviation for this year. Find the probability that the estimate is in error by more than ten hours.

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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_01\146IB_HL_OPT-Stat-Prob_01.cdr Monday, 15 April 2013 5:16:21 PM BRIAN

STATISTICS AND PROBABILITY

147

7 To work out the credit limit of a prospective credit card holder, a company gives points based on factors such as employment, income, home and car ownership, and general credit history. A statistician working for the company randomly samples 40 applicants and determines the points total for each. His results are: 14 11 13 13 15 12 12 12 10 11 11 11 12 13 14 13 11 12 14 14 14 13 15 14 11 10 11 16 11 12 12 10 11 10 10 12 13 13 13 12 a Determine the sample mean x, and standard deviation sn . b Determine a 95% confidence interval that the company would use to estimate the mean point score for the population of applicants. 8 A group of 10 students was given a revision course before their final IB examination. To determine whether the revision course was effective, the students took a test at the beginning and at the end of the course. Their marks are recorded in the table below. Student Pre-test Post-test

A 12 11

B 13 14

C 11 16

D 14 13

E 10 12

F 16 18

G 14 15

H 13 14

I 13 15

J 12 11

a Explain why it would not be appropriate to use the difference between the means of these two sets of scores to evaluate whether the revision course was effective. b Determine a 90% confidence interval for the mean of the differences in the examination scores. Explain the meaning of your answer. c It was hoped that by doing the revision course, the students’ scores would improve. Perform an appropriate test at the 5% level of significance to determine whether this was the case. 9 Yarni’s resting pulse rate was 68 beats per minute for many years. However, using a sensible diet and exercise, she hoped to reduce this rate. After six months, Yarni’s mean pulse rate is 65 beats per minute, with a standard deviation of 1:732. These statistics were calculated from 42 measurements. Using the p-value, is there sufficient evidence at a 5% level, to conclude that Yarni’s pulse rate has decreased? 10 In 2011, the mean weight of a gentoo penguin from a penguin colony was 7:82 kg with standard deviation 1:83 kg. Exactly one year after this data was found, a sample of 48 penguins from the same colony were found to have mean weight 7:55 kg. Is there sufficient evidence, at a 5% level of significance, to suggest that the mean weight in 2012 differs from that in 2011? 11

a Define the population product moment correlation coefficient ½ for the bivariate normal distribution. b Deduce that X and Y are independent random variables , ½ = 0. c The bivariate data below compares the height (x cm) and weight (y kg) of 11 men. 164 68

Height (x cm) Weight (y kg)

167 88

173 72

176 96

177 85

178 89

180 71

180 100

181 83

184 97

192 93

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i Calculate the sample product moment correlation coefficient, r. ii At a 5% level of significance, conduct a hypothesis test to determine whether or not the variables X and Y are independent. Use the p-value test only.

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148

STATISTICS AND PROBABILITY

12 Suppose distribution X has mean ¹ and variance ¾ 2 . Consider samples fx1 , x2 , x3 , ...., xn g of fixed size n > 1, of random independent values of X. ½µ n ¶ ¾ n P 2 2 1 P 1 2 Xi and Sn = Xi ¡ n X . Let X = n i=1

n

i=1

a Find Var(X). b Show that X is an unbiased estimator of ¹. c Show that Sn2 is a biased estimator of ¾ 2 . ³ ´ n Sn2 is an unbiased estimator of ¾ 2 . d Hence show that Sn2¡1 = n¡1

REVIEW SET B 1 At a country fete, there is a game where players pay to randomly choose a card from a table. There is an amount X written on the back of the card. For X being 3 or 6, the player wins $3 or $6, but for X being ¡5 or ¡1, the player has to pay an extra $5 or $1. a The i ii iii

X=x

¡5

¡1

3

P(X = x)

0:3

0:2

0:2

6

probability distribution for X is shown in the table. What is the probability of getting a 6 on card X? What is the expected return per game if the score is the return paid to the player? Explain why organisers should charge $1 to play this game, rather than 50 cents.

b At the next stall there is a similar game. It uses a Y =y ¡3 2 5 different set of cards with outcome Y which has the P(Y = y) 0:5 0:3 0:2 probability distribution shown. i Find the expected return to players for playing this game. ii Find the expected return for players wishing to play both games simultaneously. iii How much would you expect the organisers to make if people played games X and Y 500 times each, and the combined game of X and Y 1000 times, and they charged $1 for every game played? 2 A coin is biased so that when it is tossed, the probability of obtaining tails is 35 . The coin is tossed 1000 times, and X is the number of tails obtained. a Find: i the mean of X ii the standard deviation of X. b Find P(580 6 X 6 615) using X » B(100, 35 ). c Find P(579:5 6 X 6 615:5) using X » N(600, 240). d Explain why the answers in b and c are so similar. 3 By analysing the PDF of the Negative Binomial distribution, prove or disprove: “The Geometric Distribution is a special case of the Negative Binomial distribution.” 4 Suppose X » Geo(p) where x = 1, 2, 3, 4, 5, .... a Show that X has PGF given by G(t) =

pt 1 ¡ t(1 ¡ p)

for jtj
0. ¸ is a positive

ii Var(X)

c Prove that for the general exponential random variable, the median =

ln 2 . ¸

d Hence show the mean and median for the case ¸ = 0:8 on the graph in a. e Show that the cumulative density function is F (X) = P(X 6 x) = 1 ¡ e¡¸x . f Hence find P(X > 1:3) for the case ¸ = 0:8 . 6 A drink manufacturer produces soft drink for sale. Each bottle is advertised as having contents 375 mL. It is known that the machines producing these drinks are set so that the average volume dispensed into each bottle is 376 mL with standard deviation 1:84 mL. The volumes dispensed into each bottle are distributed normally. a Find the probability that an individual randomly selected bottle has volume less than 373 mL. b Find the probability that a randomly selected pack of a dozen bottles has an average volume less than the advertised amount. c Government regulations are set to ensure that companies meet their advertising claims. This company can choose to be tested under either of the following rules: I A randomly selected bottle must contain no less than 373 mL. or II A randomly selected pack of 12 bottles must have average contents no less than the advertised amount. Explain clearly by which method the company would prefer to be tested by the Government authority. d Suppose the company chose to be tested using method II above. The company wants less than 0:1% chance of being fined by the Government Authority for failing to meet the requirement. Find, to the nearest mL, what the setting should be for the average volume dispensed into each bottle, assuming the standard deviation is unchanged. 7 The manufacturer of the breakfast cereal Maxiweet knows that the net contents of each packet has variance 151:4 grams2 . A sample of 120 packets is chosen at random, and their mean weight is found to be 596:7 grams/packet. a Construct a 95% confidence interval for the true mean of the population. Interpret your answer. b The manufacturer claims that the net weight of each packet is 600 g. Is there sufficient evidence to support the manufacturer’s claim at the 5% level of significance? c Find the sample size required if the manufacturer wishes to be 95% confident that the sample mean differs from the population mean by less than 2 grams.

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8 The house prices in a large town are normally distributed. A real estate agent claims that the mean house price is E438 000. To test the real estate agent’s claim, 30 recently sold houses were randomly selected and the mean price x was calculated. The standard deviation for this sample was E23 500. What values of x, to the nearest E500, would support the agent’s claim at a 2% level of significance?

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150

STATISTICS AND PROBABILITY

9 The covariance of random variables X and Y is defined as: Cov(X, Y ) = E((X ¡ ¹X )(Y ¡ ¹Y )) a Prove that Cov(X, Y ) = E(XY ) ¡ E(X) E(Y ) b Hence, prove that if X and Y are independent, then E(XY ) = E(X) E(Y ) c Given that X and Y are independent, expand and simplify Cov(X, X ¡ Y ). 10 A machine is used to fill packets with rice. The contents of each packet weighs X grams. X is normally distributed with mean ¹ grams and standard deviation 3:71 grams. The mean weight ¹ is stated to be 500 g. To check this statement, a sample of 13 packets is selected, and the mean contents x is calculated. The hypotheses for the test are H0 : ¹ = 500 g and H1 : ¹ 6= 500 g. The critical region is defined by fx < 498g [ fx > 502g. a What is the nature of the distribution of X? b What is the meaning of the critical region? c Show that the significance level for the test is approximately 0:0519. d Find the probability of a Type II error with this test given that the true value of ¹ is 498:4 grams. 11 In 2011 a market gardener found that the weight of his tomatoes was normally distributed with mean 106:3 g and standard deviation 12:41 g. In 2012 he used a different fertiliser and found that a randomly selected sample of 65 tomatoes had mean weight 110:1 g. Assuming the population standard deviation in 2012 is again 12:41 g, is there sufficient evidence at a 1% level, to suggest that the mean weight of a tomato has increased? 12 A population X has mean ¹ and variance ¾ 2 . Random samples fx1 , x2 g of size 2 of random independent values are taken from X. a Show that T1 =

3X1 + 7X2 10

is an unbiased estimator of ¹.

b Show that T2 =

3X1 + 2X2 5

is also an unbiased estimator of ¹.

c A random sample f2:6, 5:3g is taken from X. Lucy uses T1 and Beth uses T2 to estimate ¹. Eve says that estimator T1 is more efficient than estimator T2 . i Calculate Lucy’s estimate. ii Calculate Beth’s estimate. iii Is Eve correct? Give reasons for your answer. d Let T3 =

aX1 + bX2 c

be an estimator of ¹ for constants a, b, and c 2 R + .

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State the required conditions on a, b, and c for T3 to be an unbiased estimator of ¹.

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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_01\150IB_HL_OPT-Stat-Prob_01.cdr Thursday, 18 April 2013 11:04:57 AM BRIAN

STATISTICS AND PROBABILITY

151

REVIEW SET C 1 X1 , X2 , and X3 are independent random variables where X1 » N(2, and X3 » N(a, b).

1 8 ),

X2 » N(3,

1 16 ),

a If Y = 2X3 ¡ 2X2 ¡ X1 , find: i E(Y ) ii Var(Y ) b Given that E(Y ) = 0 and Var(Y ) = 1, find the values of a and b. c State the nature of the distribution of Y . d Find P(X3 > 8b). 2 Pierre runs a game at a fair, where each player is guaranteed to win E10. Players pay a certain amount each time they roll an unbiased die, and must keep rolling until a ‘6’ occurs. When a ‘6’ occurs, Pierre gives the player E10 and the game concludes. On average, Pierre wishes to make a profit of E2 per game. How much does he need to charge for each roll of the die? 3 It is known that the probability of a journalist making no errors on each page is q. a State the distribution of the random variable X that defines the number of errors made per page by that journalist. b Find the probability, in terms of q, that the journalist makes per page: i one error ii more than one error. c The journalist gets a bonus of $10 for a page with no errors or $1 for just one error on the page, but loses $8 from their pay if there is more than one error on a page. i Draw a probability distribution table for the random variable Y which describes the returns for the journalist for making different numbers of errors on a page. ii Find E(Y ) in terms of q. iii Find the smallest value of q, 0 6 q 6 1, such that the journalist will receive an overall bonus. Give your answer to three decimal places. 4 The weekly demand for petrol (in thousands of kilolitres) at a service station is a continuous random variable with probability density function f(x) = ax3 + bx2 , 0 6 x 6 1. a If the mean weekly demand is 700 kilolitres, determine the values of a and b. b Suppose the service station has a storage capacity of 950 kilolitres. Find the probability that the service station will run out of petrol in any given week.

µ 5 The PGF for the Negative Binomial variable X is G(t) = r = 0, 1, 2, 3, 4, .... The mean of X is ¹ = a Show that p =

1¡p 1 ¡ pt

¶r for jtj
32. 8 A school claims to be able to teach anglers how to fish better and catch more fish. In order to test this hypothesis, the school recorded the number of fish caught by a random sample of nine anglers at a local jetty in a three hour period. The school then gave the anglers a free course to help them with their fishing. After the fishing course was completed, they again recorded the number of fish caught by the anglers at the same jetty in the same time period, at the same time of day. The results were: Angler

A

B

C

D

E

F

G

H

I

Number of fish caught before

24

23

22

30

41

30

33

18

15

Number of fish caught after

36

32

40

27

32

34

33

28

19

a Test at the 5% level, whether the fishing school’s claim is indeed correct. State the type of error you could make. b Find the 90% confidence interval for the mean difference of the two sets of scores, and interpret the meaning of your answer.

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9 Last year a ten-pin bowler consistently bowled a score of around 200 each game, with a standard deviation of 11:36. Her last 35 scores have had mean 196:4, and she is convinced that her form is below that of last year. Assuming her scores are normally distributed, test her claim at a 2% level, using the critical region.

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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_01\152IB_HL_OPT-Stat-Prob_01.cdr Friday, 19 April 2013 1:07:04 PM BRIAN

STATISTICS AND PROBABILITY

153

10 A random variable X has a Poisson distribution with unknown mean m. We wish to test the hypothesis H0 : m = 3 against H1 : m 6= 3. To do this, a random sample of 15 independent values fx1 , x2 , x3 , ...., x15 g is taken from X, with replacement. 15 P xi 6 56, otherwise reject H0 . The decision rule is: accept H0 if 34 6 i=1 15 P

a Define the critical region of S =

xi .

i=1

b Find the Type I error and calculate its probability. c If the true value of m is 3:4, what is the probability of a Type II error? 11 A random sample of 27 data pairs is taken from a bivariate normal distribution (X, Y ). What is the greatest value of jrj for this sample for which we can conclude that, at a 5% level of significance, X and Y are independent? Give your answer correct to 4 significant figures. 12 Suppose independent random samples of independent values are taken from the same population with unknown mean ¹ and unknown variance ¾ 2 . n 1 P For a sample of size n, the sample variance is defined by sn2 = (xi ¡ x)2 . Sample A of size 8 has sample variance

sA2

n i=1

= 4:2 .

Sample B of size 22 has sample variance sB2 = 5:1 . a Use each of the samples A and B to find two unbiased estimates of ¾ 2 . b Let t1 =

8sA2 + 22sB2 30

.

i Calculate the estimate t1 of ¾ 2 . ii Is t1 a biased or unbiased estimate of ¾ 2 ? Explain your answer. c Let t2 =

asA2 + bsB2 c

.

State the required conditions on a, b, c 2 R + for t2 to be an unbiased estimate of ¾ 2 .

REVIEW SET D 1 A soft drink manufacturer produces small and large bottles of drink. The volumes of both sizes of drink are normally distributed with means and standard deviations given in the table alongside.

small bottles large bottles

¹ (mL)

¾ (mL)

338 1010

3 12

a Find the probability that one large bottle selected at random will contain more than the combined contents of three smaller bottles selected at random. b Find the probability that one large bottle selected at random will contain three times more than one smaller bottle selected at random. 2 Patients arrive at random to the hospital Emergency room, at a rate of 14 per hour. Find the probability that: a exactly five patients arrive between 9:00 am and 9:45 am

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b fewer than seven patients will arrive between 10:00 am and 10:30 am.

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154

STATISTICS AND PROBABILITY

3 The random variable X » B(n, p) has E(X) = 8 and Var(X) = 6. a Find n and p. i P(X > 4)

b Find:

ii P(7:9 6 X 6 8:1) 1

1

¡2

4 The normal PDF with parameters ¹ and ¾ 2 is f (x) = p e 2¼¾ Z 1 1 ¡ (x¡9)2 Use f (x) to find the exact value of e 8 dx.

¡ x¡¹ ¢2 ¾

for x 2 ]¡1, 1[ .

¡1

5 The PGF for the continuous random variable X with a Gamma distribution is G(t) = (1 ¡ ¯ ln t)¡®¡1 , where ® and ¯ are parameters. a Assuming ¹ = G0 (1) and Var(X) = G00 (1) + G0 (1) ¡ [G0 (1)]2 , show that the mean of the Gamma distribution is ¹ = (® + 1)¯. b Hence determine the variance of the Gamma distribution. 6 The random variable X has a normal distribution with mean ¹. A randomly selected sample of 15 P size 15 is taken, and it is found that (xi ¡ x)2 = 230. i=1

a Find the sample variance for this sample. b Find an unbiased estimate of the population variance for the random variable X. c A confidence interval for ¹ (not the 95% confidence interval) taken from this sample is ]124:94, 129:05[ . i Find the 95% confidence interval for ¹ taken from this sample. ii Determine the confidence level for the confidence interval ]124:94, 129:05[ . 7 In order to estimate the copper content of a potential mine, drill core samples are used. All of the drill core is crushed and well mixed before samples are removed. Suppose X is the copper content in grams per kilogram of core, and that X is normally distributed with mean ¹ = 11:4 and ¾ = 9:21. Samples of size 9 are randomly chosen from the X-distribution. Histograms A and B (shown below) are for the X and X distributions, not necessarily in that order. frequency

frequency A

0

10

B

20

0

30

10

20

30

40

50

a Which of these histograms is the histogram for X? Explain your answer. b For the X-distribution, find the mean ¹X and variance ¾ X2 . c Find P(X > 12).

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d Another drill core is obtained and Y is the average copper content of n random samples. With the assumption that ¾ X = ¾ Y = 9:21, find how large n should be for Y to be within §3 grams per kg of ¹Y with 95% probability.

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155

STATISTICS AND PROBABILITY

8 A large business uses hundreds of light bulbs each year. They currently use Brand A bulbs which have a mean life of 546 hours. A supplier of another brand, Brand B, will supply bulbs at the same price as Brand A, and claims that these bulbs have a mean life in excess of 546 hours. The business will purchase Brand B bulbs if, when they test a random sample of 50 bulbs, the supplier’s claim is supported at a 5% level of significance. When the 50 bulbs were tested, the mean life was 563 hours with a variance of 3417 hours2 . Is the supplier’s claim acceptable? 9 The random variable X has a geometric distribution with unknown parameter p. Theo wishes to test the hypothesis H0 : p = 0:25 against H1 : p 6= 0:25. A random sample fx1 , x2 , x3 , ...., x12 g of independent values is taken from X, with replacement. ½ 12 ¾ ½ 12 ¾ P P xi 6 31 [ xi > 75 . The critical region is defined as a If S =

12 P

i=1

i=1

xi , what is the distribution of S under H0 ?

i=1

b What is the acceptance region for S under the null hypothesis H0 ? c Calculate the level of significance ® for this test. d The true value of p is p = 0:2 . Calculate the power of the test. 10 Quickchick grow chickens to sell to a supermarket chain. However, the buyers believe that the chickens are supplied underweight. As a consequence they consider the hypotheses: H0 : Quickchick is not supplying underweight chickens H1 : Quickchick is supplying underweight chickens. a a type I error

What conclusion would result in:

b a type II error?

11 An archer shoots 10 arrows at a target from each of 12 different positions. The table below shows the distance of each position from the target, and how many shots were successful. Position Distance from target (x m)

A 20

B 25

C 15

D 35

E 40

F 55

G 30

H 45

I 60

J 80

K 65

L 70

Hits (y)

9

8

8

8

7

6

9

7

4

2

3

3

a Draw a scatter plot of the data. b Calculate the sample correlation coefficient r. c Comment on the strength of the correlation between X and Y . d Is there a causal relationship between the variables? e Find the equation of the regression line of Y on X. f Predict the number of hits made from 50 m if the archer shoots 100 arrows. g Should the regression line be used to predict the number of hits when arrows are fired from 100 m? Explain your answer.

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h Find the equation of the regression line of X on Y .

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156

STATISTICS AND PROBABILITY

12 X is a random variable with unknown mean ¹ and unknown variance ¾ 2 . Consider samples of size 2, fx1 , x2 g, of independent values taken from X. Let T1 =

2X1 + X2 3

and T2 = aX1 + (1 ¡ a)X2

for some constant a 2 [0, 1].

a For each estimator T1 and T2 , determine whether it is a biased or unbiased estimator of ¹. b Calculate Var(T1 ) and Var(T2 ). c For which value(s) of a is T2 a more efficient estimator than T1 ? d For which value of a 2 [0, 1] is the estimator T2 most efficient? 13 Suppose X and Y are independent random variables with E(X) = ¹X , Var(X) = ¾ X2 , E(Y ) = ¹Y , and Var(Y ) = ¾ Y2 . n P 2 A random sample of size n is taken from X, and the sample mean x and

sX2

=

are calculated.

(xi ¡ x)

i=1

n¡1 m P

Similarly, a sample of size m is taken from Y , and the sample mean y and

sY2

=

(yi ¡ y)2

i=1

m¡1

are calculated. Let U = 3X ¡ 5Y . a Find E(U ) and Var(U ) in terms of ¹X , ¹Y , ¾ X2 , and ¾ Y2 where appropriate.

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STATISTICS AND PROBABILITY

THEORY OF KNOWLEDGE

157

THE CENTRAL LIMIT THEOREM

The French mathematician Abraham de Moivre is most famous for his formula linking complex numbers and trigonometry, but he was also responsible for the early development of the Central Limit Theorem. In an article from 1733, he used the normal distribution to approximate the distribution of the number of heads resulting from many tosses of a fair coin. The work of de Moivre was extended in 1812 by country-man Pierre-Simon Laplace, but the theorem was not formalised and rigorously proven until the early 20th century work of the Russian mathematicians Pafnuty Chebyshev, Andrey Markov, and Aleksandr Lyapunov. In 1889, Sir Francis Galton wrote about the normal distribution, with particular relevance to the Central Limit Theorem:

Abraham de Moivre

“I know of scarcely anything so apt to impress the imagination as the wonderful form of cosmic order expressed by the law of frequency of error. The law would have been personified by the Greeks if they had known of it. It reigns with serenity and complete self-effacement amidst the wildest confusion. The larger the mob, the greater the apparent anarchy, the more perfect is its sway. It is the supreme law of unreason.” 1 How is mathematical order linked to our sense of beauty? The name “Central Limit Theorem” was first used in 1920 by the Hungarian mathematician George P´olya (1887-1985). He used the word “central” because of the importance of the theorem in probability theory. 2 What makes a theorem “fundamental”? Could the Central Limit Theorem reasonably be referred to as the Fundamental Theorem of Statistics? 3 Discuss the statement: “Without the Central Limit Theorem, there could be no statistics of any value within the human sciences.” The philosophies of rationalism and empiricism concern the extent to which we are dependent upon sense experience in order to gain knowledge. Rationalists claim that knowledge can be gained through logic independent of sense experience. They argue about the limitations of what sense experience can provide, and how reason in other forms contributes additional information about the world. By contrast, empiricists claim that sense experience is most important for knowledge, arguing that we cannot regard something as knowledge if experience cannot provide it.

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158

STATISTICS AND PROBABILITY

THEORY OF KNOWLEDGE

POPULATION PARAMETERS

In previous courses we have seen that data collected from a population may be qualitative or quantitative. However, there are other properties of parameters that can affect how we need treat them. We may need to sample for data in a particular way, or word questions carefully so the person responding understands the context in which the question is being asked. Data is objective if the result of its measurement is clear. For example: ² The number of cousins you have is objective, quantitative, and discrete. ² A person’s height, and weight are all objective, quantitative, and continuous. ² Either a person is infected with HIV, or they are not, so this parameter is objective and qualitative. By contrast, data is subjective if the result of its measurement depends on individual interpretation, and is relative to the individual’s own experiences. For example: ² The colour of a person’s shirt is subjective and qualitative. For some colours like black and white, people will agree on the result, but people may argue about whether a shirt is red or pink, or perhaps blue or purple. ² A person’s mood is subjective and qualitative. We can ask a person to rate themselves on a scale of sad to happy, or confident to fearful, but some states such as angry do not necessarily have defined opposites. Moods are also transitive in the sense that they can change rapidly, so a data set correct at the time of measurement may be very different to that which would be measured from the same sample of people soon after. 1 Is subjective data just the product of a vague question? For example, can the question “How many friends do you have?” be refined so it is no longer subjective? 2 Are statistics gathered on transitive parameters less meaningful? 3 What do you regard as the most important things in life? Are these things qualitative or quantitative? Are they objective or subjective? Are they transitive? 4 To what extent is “well-being” a social construct? 5 What sort of social consciousness is necessary for a subjective response to fit into the broader context of society? 6 Does the ability to test only certain parameters in a population affect the way knowledge claims in the human sciences are valued? The measurement of performance poses particular challenges for statisticians because a contextual framework needs to be set in which the performance is measured.

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For example, the performance of an athlete may be measured by how fast they run, how high they jump, or how far they throw, but in doing this we require a standard for comparison. Do we compare the results of an athlete to those of other athletes of the same age or the same gender or the same race

IB HL OPT 2ed

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STATISTICS AND PROBABILITY

159

or the same body physique, or against all athletes in the world? Or is it more important to compare the athlete’s results against themselves and what is physically achieveable for the individual? 7 How can we measure evolutionary success? You may wish to think about number of progeny, life-span of a species, and diversity within a species.

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8 How can we measure the performance of an organisation? Should the performance be measured by its owners, its managers, its employees, or its customers? How does profit balance against service, or company mission?

IB HL OPT 2ed

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160

WORKED SOLUTIONS 4

a

Worked Solutions ¾2 = 22 p ) ¾ = 22

1 X » N(5, 22).

1

2

1 4

1 2

1 4

x2

0

1

4

Probability

1 4

1 2

1 4

b P(X = 30) = P(29:5 6 X < 30:5) ¼ 0:000 000 060 8

=

1 3

Z

5

0

·

1 6

y2

1

4

9

16

25

32

Probability

1 6

1 6

1 6

1 6

1 6

1 6

x+y

1

2

3

4

5

6

7

8

Probability

1 24

1 8

1 6

1 6

1 6

1 6

1 8

1 24

P(X + Y = 5) or X = 2, Y = 3)

y

1 5

x2 2

£

= 2:5

=

¸5

4

9

16

25

36

49

64

Probability

1 24

1 8

1 6

1 6

1 6

1 6

1 8

1 24

4x ¡ 2y

¡12 ¡10 ¡8 ¡6 ¡4 ¡2 0 2 4

Probability

0

25 2

¡3

¡1

2

3

P(X = x)

1 6

1 3

1 3

1 6

P

1

iv

x

5

1 24

=

1 3

+

2 3

+

b

64

100

144

Probability

1 6

1 8

1 24

1 24

P((4X ¡ 2Y )2 = 4)

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0

1

2

3

4

5

6

8

10

12

Prob.

1 4

1 12

1 8

1 12

1 8

1 12

1 8

1 24

1 24

1 24

(xy)2 0

1

4

9

1 4

1 12

1 8

1 12

1 8

1 12

1 8

1 24

1 24

1 24

E(X) = 0( 14 ) + 1( 12 ) + 2( 14 ) = 1 = 0( 14 ) + 1( 12 ) + 4( 14 ) ¡ 12 =

1 9

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1 2

E(Y ) = 1( 16 ) + 2( 16 ) + 3( 16 ) + 4( 16 ) + 5( 16 ) + 6( 16 ) = 3:5 Var(Y ) = E(Y 2 ) ¡ [E(Y )]2 = 1( 16 ) + 4( 16 ) + 9( 16 ) + 16( 16 ) + 25( 16 )

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+ 36( 16 ) ¡ 3:52 ¼ 2:92

5

95

16 25 36 64 100 144

Var(X) = E(X 2 ) ¡ [E(X)]2

100

50

75

25

0

1 + 16 8 7 24

xy

i

ii

and (E(X))2 =

5

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0

5

14 3

1 24

36

Prob.

xi2 pi

E(X 2 ) 6= (E(X))2

1 24

5 24

v

= 1( 13 ) + 4( 13 ) + 9( 13 )

)

1 8

16

=

P(X 2 = 9) = P(X = 3 or ¡3) = 13

b E(X 2 ) =

1 8

7 24

=

1 2

14 3

1 6

= P(4X ¡ 2Y = 2 or ¡2)

P(X = 4) = P(X = 2 or ¡2) = P(X = 2) = 13

=

1 6

4

2

) E(X 2 ) =

1 8

1 8

For example:

Possible X 2 values are 1, 4, and 9. P(X 2 = 1) = P(X = ¡1 or 1) = P(X = ¡1) = 13

P

1 8

0

= ¡3( 16 ) ¡ 1( 13 ) + 2( 13 ) + 3( 16 ) 1 3

1 24

6

(4x ¡ 2y)2

xi pi

= ¡ 12 ¡

1 6

(x + y)2

y = f(x)

Qt

x

a E(X) =

6

1 6

= P(X = 0, Y = 5 or X = 1, Y = 4,

x( 15 ) dx

=

1 5

5

1 6

= ( 14 )( 16 ) + ( 12 )( 16 ) + ( 14 )( 16 ) x f (x) dx

0

=

4

1 6

5

E(X) =

=

3

1 6

1 2

+

b X » U(0, 5)

Z

2

1 6

For example:

a E(X) = ¡1( 12 ) + 1( 13 ) + 3( 16 ) 1 3

1

iii

c P(X = 11) = P(10:5 6 X < 11:5) ¼ 0:0376

= ¡ 12 +

y Probability

ii

a P(X = 24) = P(23:5 6 X < 24:5) ¼ 0:000 023 9

3

0

For example: P(X 2 = 4) = P(X = 2 or ¡2) = 14 + 0 = 14

EXERCISE A

2

x Probability

i

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IB HL OPT 2ed Calculus

WORKED SOLUTIONS 1 E(X + Y ) = 1( 24 ) + 2( 18 ) + 3( 16 ) + 4( 16 ) + 5( 16 )

iii

+

6( 16 )

+

7( 18 )

+

d E(5XY ) = 5E(XY ) =5£6 = 30

1 8( 24 )

= 4:5

If X and Y are dependent, E(X + Y ) = 5. However, E(XY ), Var(2X ¡ 3Y + 6), and E(5XY ) can only be determined if X and Y are independent.

Var(X + Y ) = E((X + Y )2 ) ¡ [E(X + Y )]2 1 = 1( 24 ) + 4( 18 ) + 9( 16 ) + 16( 16 ) + 25( 16 ) 1 + 36( 16 ) + 49( 18 ) + 64( 24 ) ¡ 4:52

6

¼ 3:42

x

0

1

2

3

Probability

1 4

1 4

1 4

1 4

a

E(4X ¡ 2Y )

iv

E(X) = 0( 14 ) + 1( 14 ) + 2( 14 ) + 3( 14 )

1 1 = ¡12( 24 ) ¡ 10( 24 ) ¡ 8( 18 ) ¡ 6( 18 ) ¡ 4( 16 )

) E(X) =

1 1 ) + 6( 24 ) ¡ 2( 16 ) + 0( 18 ) + 2( 18 ) + 4( 24

Var(4X ¡ 2Y ) = E((4X ¡ 2Y )2 ) ¡ [E(4X ¡ 2Y )]2

0

1

4

9

Probability

1 4

1 4

1 4

1 4

) E(X 2 ) = 0( 14 ) + 1( 14 ) + 4( 14 ) + 9( 14 )

7 5 = 0( 18 ) + 4( 24 ) + 16( 24 ) + 36( 16 ) + 64( 18 )

) E(Y ) =

1 1 + 100( 24 ) + 144( 24 ) ¡ (¡3)2

¼ 19:7

E(XY ) 1 1 1 + 6( 18 ) + 8( 24 ) + 10( 24 ) + 12( 24 )

= 3:5 Var(XY )

xy = x3

0

1

8

27

Probability

1 4

1 4

1 4

1 4

x

¡1

0

1

2

Probability

1 4

1 4

1 4

1 4

b 2

= E((XY ) ) ¡ [E(XY )]

1 1 1 = 0( 14 ) + 1( 12 ) + 4( 18 ) + 9( 12 ) + 16( 18 ) + 25( 12 )

E(X) = ¡1( 14 ) + 1( 14 ) + 2( 14 )

1 1 1 ) + 100( 24 ) + 144( 24 ) ¡ 3:52 + 36( 18 ) + 64( 24

) E(X) =

= 10:5

ii

iii

E(X + Y ) = 4:5 E(X) + E(Y ) = 1 + 3:5 = 4:5 X Var(X + Y ) ¼ 3:42 Var(X) + Var(Y ) ¼ 12 + 2:917 ¼ 3:42 X

Var(Y ) =

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1 4

xy = x3

¡1

0

1

8

Probability

1 4

1 4

1 4

1 4

E(X)E(Y ) =

3 4

6= E(XY )

a E(X) = 3:8, E(Y ) = 5:7 E(3X ¡ 2Y ) = 3E(X) ¡ 2E(Y ) = 3 £ 3:8 ¡ 2 £ 5:7 =0 Var(3X ¡ 2Y ) = 9Var(X) + 4Var(Y ) = 9 £ 0:3232 + 4 £ 1:022 = 5:100 561 p ) the standard deviation of 3X¡2Y is 5:100 561 which is ¼ 2:26 . b By Theorem 8, the linear combination 3X ¡ 2Y of independent normally distributed random variables X and Y is also normally distributed ) 3X ¡ 2Y » N(0, 2:258 442 ) Thus P(3X ¡ 2Y > 3) ¼ 0:0920 fusing technologyg

E(XY ) = E(X)E(Y ) =3£2 =6

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b

75

25

0

5

95

Var(2X ¡ 3Y + 6) = 4Var(X) + 9Var(Y ) + Var(6) = 4( 32 ) + 9( 54 ) + 0 = 17:25

100

50

c

75

25

0

5

E(X + Y ) = E(X) + E(Y ) =3+2 =5

4

1 2

)

If X and Y are independent, a

1

1 4

) E(XY ) = ¡1( 14 ) + 1( 14 ) + 8( 14 ) =2

E(XY ) = 3:5 E(X)E(Y ) = 1 £ 3:5 = 3:5 X

3 , 2

0

Probability

) E(Y ) = 1 12

E(4X ¡ 2Y ) = ¡3 4E(X) ¡ 2E(Y ) = 4 £ 1 ¡ 2 £ 3:5 =4¡7 = ¡3 X Var(4X ¡ 2Y ) ¼ 19:7 16Var(X) + 4Var(Y ) ¼ 16( 12 ) + 4(2:917) ¼ 19:7 X

5 E(X) = 3, E(Y ) = 2, Var(X) =

y = x2

E(Y ) = 1( 12 ) + 4( 14 )

95

i

1 2

100

c

7 2

) E(XY ) = 0( 14 ) + 1( 14 ) + 8( 14 ) + 27( 14 ) =9 E(X)E(Y ) = 32 £ 72 = 21 6= E(XY ) 4

1 1 1 ) + 2( 18 ) + 3( 12 ) + 4( 18 ) + 5( 12 ) = 0( 14 ) + 1( 12

2

3 2

y = x2

= ¡3

v

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IB HL OPT 2ed Calculus

162

WORKED SOLUTIONS

8 X » N(¹, ¾ 2 )

12 S » N(280, 4) and L » N(575, 16) a We need to find P(L < 2S), which is P(L ¡ 2S < 0). If D = L ¡ 2S, E(D) = E(L) ¡ 2E(S) = 575 ¡ 2 £ 280 = 15

Now P(X > 80) = 0:1 and P(X > 65) = 0:3 ) P(X < 80) = 0:9 and P(X < 65) = 0:7

´

80 ¡ ¹ = 0:9 and ¾ ¾ ³ X ¡ ¹ 65 ¡ ¹ ´ < = 0:7 P ¾ ¾ ³ ´ ³ ´ 80 ¡ ¹ 65 ¡ ¹ ) P Z< = 0:9 and P Z < = 0:7 ¾ ¾ 80 ¡ ¹ 65 ¡ ¹ ¼ 1:281 55 and ¼ 0:524 40 ) ¾ ¾
5S) = P(L ¡ 5S > 0) Now E(L ¡ 5S) = E(L) ¡ 5E(S) = 90 ¡ 5 £ 21 = ¡15 and Var(L ¡ 5S) = Var(L) + 25Var(S) = 15 + 25 £ 5 = 140 ) L ¡ 5S » N(¡15, 140) and P(L ¡ 5S > 0) ¼ 0:102 fusing technologyg

13

) P(S > 440) ¼ 0:888 Assumption: The random variables A1 , A2 , A3 , A4 , C1 , C2 , and C3 are independent.

10 Let C be the amount of black coffee dispensed and let F be the amount of froth. Then C » N(120, 72 ) and F » N(28, 4:52 ) Consider S = C + F E(S) = E(C) + E(F ) = 120 + 28 = 148 mL

P(L > S1 + S2 + S3 + S4 + S5 ) = P(L ¡ S1 ¡ S2 ¡ S3 ¡ S4 ¡ S5 > 0) Now E(L ¡ S1 ¡ S2 ¡ S3 ¡ S4 ¡ S5 ) = E(L) ¡ E(S1 ) ¡ E(S2 ) ¡ E(S3 ) ¡ E(S4 ) ¡ E(S5 ) = 90 ¡ 21 ¡ 21 ¡ 21 ¡ 21 ¡ 21 = ¡15 and Var(L ¡ S1 ¡ S2 ¡ S3 ¡ S4 ¡ S5 ) = Var(L) + Var(S1 ) + Var(S2 ) + Var(S3 ) + Var(S4 ) + Var(S5 ) = 15 + 5 £ 5 = 40 ) L ¡ S1 ¡ S2 ¡ S3 ¡ S4 ¡ S5 » N(¡15, 40) and P(L ¡ S1 ¡ S2 ¡ S3 ¡ S4 ¡ S5 > 0) ¼ 0:008 85

b We need to find

Var(S) = Var(C) + Var(F ) = 49 + 4:52 = 69:25 mL2 P(U < 135:5) ¼ 0:0665 ¼ 6:65% which is greater than 1% ) proprietor needs to adjust the machine.

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P

a As P (x) = 1 in each distribution, each is a well defined probability distribution. b ¹X = E(X) P = x P (x) = ¡3(0:25) ¡ 2(0:25) + 3(0:25) + 5(0:25) = 0:75

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25

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11 X » N(¡10, 1) and Y » N(25, 25) a E(U ) = E(3X + 2Y ) = 3E(X) + 2E(Y ) = 3(¡10) + 2(25) = 20 Var(U ) = 9Var(X) + 4Var(Y ) = 9(1) + 4(25) = 109 ) standard deviation of U ¼ 10:4 b P(U < 0) ¼ 0:0277 fusing technologyg

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fusing technologyg

b Now we need to find P(L < S1 + S2 ) = P(L ¡ S1 ¡ S2 < 0) where E(L ¡ S1 ¡ S2 ) = E(L) ¡ E(S1 ) ¡ E(S2 ) = 575 ¡ 280 ¡ 280 = 15 and Var(L ¡ S1 ¡ S2 ) = Var(L) + Var(S1 ) + Var(S2 ) = 16 + 4 + 4 = 24 Thus L ¡ S1 ¡ S2 » N(15, 24) and P(L ¡ S1 ¡ S2 < 0) ¼ 0:001 10 fusing technologyg

100

³X ¡ ¹

) P

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IB HL OPT 2ed Calculus

WORKED SOLUTIONS EXERCISE B.1

Var(X) = E(X 2 ) ¡ [E(X)]2

1

= 9(0:25) + 4(0:25) + 9(0:25) + 25(0:25) ¡ 0:752 = 47 £ 0:25 ¡ 0:752 = 11:1875 and so ¾ X ¼ 3:34 ¹Y = E(Y ) = ¡3(0:5) + 2(0:3) + 5(0:2) = 0:1

5

2

3

8

10

(0.3)

2

-1

0

5

7

(0.5) -3

-6

-5

0

2

20 k

25 k

30 k

x

5

10

15

20

25

30

P(X = x)

1 6

1 6

1 6

1 6

1 6

1 6

b ¹ = E(X) = 5( 16 ) + 10( 16 ) + 15( 16 ) + :::: + 30( 16 ) = 17:5 c P(X < ¹) = P(X < 17:5)

X + Y values

(0.2)

15 k

The probability distribution is:

c With X, the expected win is $0:75/game. However, it costs $1 to play each game, so there is an expected loss of $0:25/game. With Y there is an expected loss of $0:90/game. d As ¾ X > ¾ Y we expect greater variation in the results of game X. Y

10 k

6k = 1 k = 16

)

= 9(0:5) + 4(0:3) + 25(0:2) ¡ 0:12 = 10:69 and so ¾ Y ¼ 3:27

i

5 k

x P(X = x)

a

Var(Y ) = E(Y 2 ) ¡ [E(Y )]2

e

163

= P(X = 5, 10, or 15) 1 6 1 2

= =

+

1 6

1 6

+

d Var(X) = E(X 2 ) ¡ [E(X)]2 = 25( 16 ) + 100( 16 ) + 225( 16 ) + 400( 16 ) + 625( 16 ) + 900( 16 ) ¡ 17:52

= 72:916 66::::

P(¡6) = 0:25 £ 0:5 = 0:125 P(¡1) = 0:25 £ 0:3 = 0:075

X

) ¾=

P(¡5) = 0:25 £ 0:5 = 0:125 P(0) = 0:25 £ 0:5 + 0:25 £ 0:3 = 0:200 P(3) = 0:25 £ 0:2 = 0:050

P(2) = 0:25 £ 0:2 + 0:25 £ 0:5 = 0:175 P(5) = 0:25 £ 0:3 = 0:075 P(8) = 0:25 £ 0:2 = 0:050

) F (0) =

¡5 0:125

¡1 0:075

0 0:2

2 0:175

X +Y P(X + Y )

3 0:05

5 0:075

7 0:075

8 0:05

10 0:05

P x

3 ) 4

pk (1 ¡ p)1¡k

k=0 ( 34 )0 ( 14 )1 ( 34 )0 ( 14 )1

=

1 4 ( 34 )1 ( 14 )0

3 X » B(7, p) where p < 0:5 . P(X = 4) =

¡7¢ 4

p4 (1 ¡ p)3 = 0:097 24

) 35p4 (1 ¡ p)3 = 0:097 24 ) p4 (1 ¡ p)3 ¼ 0:002 778 29 ) p ¼ 0:299 999 43 or 0:815 481 36 But p < 0:5, so p ¼ 0:300 and P(X = 2) =

¡7¢ 2

(0:3)2 (0:7)5

¼ 0:318 4 p = 0:35 is the probability of rain on an August day. X » B(7, 0:35) a P(X = 3) =

¡7¢ 3

(0:35)3 (0:65)4

¼ 0:268 b P(X > 3) = 1 ¡ P(X 6 2) ¼ 0:468

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c P(X 6 3) ¼ 0:800

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=1¡p

F (1) = + =1 Interpretation: F (0) is the probability of ‘no reds’ F (1) is the probability of ‘at most one red’ ‘at most one red’ is 0 reds or 1 red which is a certain event.

+ :::: + 100(0:05) ¡ (0:85)2 = 21:8775 ) ¾ U ¼ 4:68 iii With the new game there is an expected loss of $1 ¡ $0:85 = $0:15/game.

0

F (x) =

b

ii U = X + Y E(U) = ¡6(0:125) ¡ 5(0:125) ¡ 1(0:075) + :::: + 10(0:05) = 0:85 ) ¹U = 0:85 Var(U) = 36(0:125) + 25(0:125) + 1(0:075)

5

=p

) X » B(1,

¡6 0:125

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3 4 1 4

P(X = 0) =

P(7) = 0:25 £ 0:3 = 0:075 P(10) = 0:25 £ 0:2 = 0:050

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Var(X) ¼ 8:54

a P(X = 1) =

2

X +Y P(X + Y )

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p

100

5 -3 3 -2 (0.25) (0.25) (0.25) (0.25)

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IB HL OPT 2ed Calculus

164

WORKED SOLUTIONS d Days of rain:

D2 2nd 3rd 4th 5th 6th

D3 3rd 4th 5th 6th 7th

9 > > = > > ;

8

L

5

0.4 C 0.2

P(rains on exactly 3 days in succession)

0.25

= 5 £ (0:35)3 (0:65)4 ¼ 0:0383 Assumptions: Rain falling on any day is independent of rain falling on any other day.

Is 0.15 In

Now P(X > 1) > 0:9 ) P(X = 0) < 0:1 0

) (0:8)n < 0:1 ) n log(0:8) < log(0:1)

=

(0:25)(0:7) (0:4)(0:7) + (0:2)(0:45) + (0:25)(0:7) + (0:15)(0:4)

)

b Let X = number of multiples of 7 obtained in 500 spins. Then X » B(500, 0:14) Now 15% of 500 = 75 and P(X > 75) = 1 ¡ P(X 6 75) ¼ 0:237 fusing technologyg c

n log(0:7) 6 log(0:02)

log(0:02) log(0:7)

) n > 10:968:::: least n is n = 11.

7 Let X = the number of letters addressed to the Accounts Department. X » B(20, 0:7) a P(X > 11) = 1 ¡ P(X 6 10) ¼ 0:952

=

EXERCISE B.2 1 X » Geo(0:25)

p p

a

np(1 ¡ p)

70 £ 0:7 £ 0:3

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25

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5

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25

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50

75

25

0

5

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100

50

75

25

0

5

P(X = 4) = p(1 ¡ p)3 = 0:25 £ (0:75)3 ¼ 0:105

¼ 3:83

cyan

i E(X) = np = 500 £ 0:14 = 70 ) the school expects to make in 500 spins (500 ¡ 70) £ $20 ¡ 70 £ $100 = $1600 ii The school loses if (500 ¡ X)20 ¡ 100X < 0 ) 500 ¡ X ¡ 5X < 0 ) 6X > 500 ) X > 83 13 and P(X > 83 13 ) = 1 ¡ P(X 6 83) ¼ 0:0435

b Y » B(70, 0:7) ¹ = np = 70 £ 0:7 = 49 letters ¾=

fBayes Theoremg

a There are 7 multiples of 7 which are < 50. 7 ) P(multiple of 7) = 50 = 0:14

9

(0:7)n 6 0:02

n>

S

P(P j Is)P(Is) P(P)

c We need to find the smallest n such that 1 ¡ (0:7)n > 0:98

)

P X

0.6

¼ 0:417 72 If X is the number of parcels selected then X » B(2, 0:417 72) and P(X = 1) ¼ 0:486. Assumption: The events are independent.

P(still operating with n cells) = P(X 6 n ¡ 1) = 1 ¡ P(X = n) = 1 ¡ (0:7)n

)

0.4

=

ii P(still operating) = P(X 6 14) ¼ 1 ¡ 0:004 75 ¼ 0:995 25 ¼ 0:995

)

S

P(S j leaves state)P(leaves state) P(S) (0:25)(0:3) + (0:15)(0:6) = (0:4)(0:3) + (0:2)(0:55) + (0:25)(0:3) + (0:15)(0:6)

log(0:1) = 10:318:::: log(0:8)

a Let X be the number of cells failing in one year. X » B(15, 0:7) i P(X = 15) ¼ 0:004 75

b

P

0.7 0.3

b For one parcel, P(leaves state j S)

) least n is n = 11. We are assuming the binomial model even though it is not strictly binomial. 6

0.55 S

¼ 0:289

(0:2)0 (0:8)n < 0:1

) n>

0.45 P X

=

95

¡n¢

S

a P(Is j P)

5 Let X be the number of red pens selected. Due to the very large number of pens, the number of reds selected in n attempts is approximately X » B(n, 0:2)

)

P X

0.7 0.3

b

P(X 6 2) = P(X = 1 or 2) = p + p(1 ¡ p) = 0:4375

100

)

D1 1st 2nd 3rd 4th 5th

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\164IB_HL_OPT-Stat-Prob_an.cdr Wednesday, 17 April 2013 1:39:52 PM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS c

P(X > 3) = 1 ¡ P(X 6 3) = 1 ¡ P(X = 1, 2, or 3) = 1 ¡ [p + p(1 ¡ p) + p(1 ¡ p)2 ] ¼ 0:422

b

2 X » Geo(p), then P(X = x) = p(1 ¡ p)x¡1 for x = 1, 2, 3, 4, 5, .... . Now

P 1

=p

1 1 ¡ (1 ¡ p)

8

¸

a

· ¸

)

1 p

3

=1 the probability distribution is well defined.

¡6¢ 2

(0:29)3 (0:71)4

1

a P(X = x) =

)

a X = number of games needed for Eva to win 3 games Then X » NB(3, 0:35)

2

(0:35) (0:65)

2

a X » Geo(0:72) ) P(X = 5) ¼ 0:004 43 b Y » NB(4, 0:72)

¡ 11 ¢ 3

2

(0:72)4 (0:28)8

i

yellow

2 p 6 ¼ 4:449 48::::

Since 2Var(X) = [E(X)]2 ¡ 15, 2m = m2 ¡ 15

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

50

75

25

0

5

95

100

50

75

25

0

100

magenta

p 6

16 ¡ 4(1)(¡2)

) m2 ¡ 2m ¡ 15 = 0 ) (m + 3)(m ¡ 5) = 0 ) m=5 fas m > 0g ) the mean is 5 calls/hour.

7 X » Geo(0:15) a P(first snow on Nov 15) = P(X = 15) ¼ 0:0154

5

5 6

a X is a Poisson random variable as the average number of phone calls to the police per hour is constant. We assume that the average number of calls each hour is constant. b

¼ 0:001 68

cyan

=

p

But m > 0, so m = 2 + ) ¹ ¼ 4:45 b P(1 6 X 6 5) = P(X 6 5) ¡ P(X = 0) ¼ 0:711 53 ¡ 0:011 68 ¼ 0:700

(0:35) (0:65)

) P(X = 12) =

4§

) m=2§

¼ 0:235 6

30 36

m2 e¡m e¡m 2me¡m = + 2! 0! 1! m2 ) = 1 + 2m 2

) m=

¡2¢ ¡ ¢ (0:35)3 (0:65)0 + 32 (0:35)3 (0:65)1 2 ¡4¢ 3 2 +

0 1 2 3 4 5 6

for x = 0, 1, 2, 3, 4, ....

) m ¡ 4m ¡ 2 = 0

1

P(Eva beats Paul in a match) = P(X = 3, 4, or 5) =

mx e¡m x!

2

¼ 0:0836 c

1 0 1 2 3 4 5

) m2 = 4m + 2

P(Eva beats Paul 3 games to 1) = P(X = 4) =

2 1 0 1 2 3 4 1

P(X = 2) = P(X = 0) + 2P(X = 1)

and P(X > 3) = 1 ¡ P(X 6 2) = 0:0289 fusing technologyg

3

3 2 1 0 1 2 3 Die

EXERCISE B.3

P(X = 3) = p(1 ¡ p)2 = 0:023 987 Solving using technology gives p = 0:83 fas p > 0:5g

¡3¢

4 3 2 1 0 1 2

b X » ) P(player 1 is first to start on 2nd roll) = P(X = 5) fall 4 players fail on 1st attemptg ¼ 0:000 643

4 X » Geo(p)

b

5 4 3 2 1 0 1

G( 56 )

¼ 0:0930

5

f log 0:85 < 0g

P(difference is no more than 3) =

a X » Geo(0:29) Using technology, ) P(X = 4) ¼ 0:104 b Y » NB(3, 0:29) ) P(X = 7) =

log(0:15) log(0:85)

Difference table 6 5 4 3 2 1

Die 2

fas 1 + (1 ¡ p) + (1 ¡ p)2 + :::: is an infinite GS with u1 = 1 and r = 1 ¡ p with 0 < r < 1.g =p

n>

) n > 11:673:::: least n is n = 12 must book for Dec 12.

) )

= p + p(1 ¡ p) + p(1 ¡ p)2 + p(1 ¡ p)3 + :::: = p[1 + (1 ¡ p) + (1 ¡ p)2 + (1 ¡ p)3 + ::::]

·

P(snow falls on or before n days) = 1 ¡ P(snow does not fall in n days) = 1 ¡ (0:85)n So, we need to solve 1 ¡ (0:85)n > 0:85 ) (0:85)n < 0:15 Thus, n log(0:85) < log(0:15) )

P(X = i) = P(X = 1) + P(X = 2) + P(X = 3) + ::::

i=1

165

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\165IB_HL_OPT-Stat-Prob_an.cdr Tuesday, 16 April 2013 5:45:54 PM BRIAN

IB HL OPT 2ed Calculus

166

WORKED SOLUTIONS ii Thus X » Po(5) and P(X 6 3) ¼ 0:265. 50 i The probability of a fault in a 50 m length = 1000 = 0:05 ) X » Po(0:05) ) P(X = 0) ¼ 0:951

a

ii

fBinomial theoremg Thus, X + Y is Poisson with mean m1 + m2 . b Pn is “If X1 , X2 , X3 , ...., Xn are independent Poison random variables with corresponding means m1 , m2 , m3 , ...., mn then X1 + X2 + :::: + Xn is a Poisson random variable with mean m1 + m2 + m3 + :::: + mn ” Proof by induction:

P(at most 2 faults in 50 m) = P(X 6 2) ¼ 0:999 98 which is ¼ 1

If n = 1, P(X = x) =

a Let X = number of internal calls, and Y = number of external calls

) P1 is true. If Pj is true then X1 + X2 + :::: + Xj has P(X1 + X2 + :::: + Xj = l) has PDF

), for 5 minutes. ) X » Po( 54 ) and Y » Po( 10 6 The total number of calls each five minute session is X + Y where E(X + Y ) = E(X) + E(Y ) = 54 + 10 6 ¼ 2:917 and Var(X + Y ) = Var(X) + Var(Y ) ¼ 2:917 also ) X + Y » Po(2:917) fassuming X and Y are independent random variablesg ) P(X + Y = 3) ¼ 0:224 b As E(X + Y ) ¼ 2:917, the receptionist should expect 3 calls each 5 minutes. c i P(X + Y > 5) = 1 ¡ P(X + Y 6 5) ¼ 0:0758 ii

5 calls in 20 min = x calls in 7 min ) x = 74 10 calls in 30 min = y calls in 7 min ) y = 73 )

E(X + Y ) = Var(X + Y ) =

7 4

+

l P

i=0

=

l P

P(X = x) = P(Y = y) =

x!

=

k P k P

¼ 4:083

a ¹=

1

and

1 p

q p2 0:667 = (0:333)2

b Var(X) =

1 = 0:333 ¼ 3:00

¼ 6:015:::: ) ¾ ¼ 2:45 a X » Geo(0:25)

2

1 ) ¹= p

b Y » NB(2, 0:25) ) ¹=

1 = 0:25 = 4 throws

r p

2 = 0:25 = 8 throws

3 X » Geo(0:05)

i!

£

a ¹=

m2k¡i e¡m2

1 p

1 = 0:05 = 20 throws

(k ¡ i)!

magenta

yellow

50

75

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

l! i!(l ¡ i)!

EXERCISE B.4

[P(X = i) £ P(Y = k ¡ i)]

5

95

l¡i (m1 + m2 + :::: + mj )i £ mj+1 £

1 e¡m1 ¡m2 P k! m1i m2k¡i k! i!(k ¡ i)! i=0

100

50

75

25

0

5

(l ¡ i)!

y!

1 m i e¡m1 P 1

cyan

i!

£ (m1 + m2 + :::: + mj + mj+1 )l Thus Pj+1 is true whenever Pj is true, and P1 is true. ) Pn is true. fPrinciple of mathematical inductiong

[P(X = i and Y = k ¡ i)]

i=0

=

l¡i ¡mj+1 mj+1 e

#

e¡(m1 +m2 +::::+mj +mj+1 ) £ l!

y m2 e¡m2

i=0

=

(m1 + m2 + :::: + mj )i e¡(m1 +m2 +::::+mj )

i=0

i=0

=

"

e¡(m1 +m2 +::::+mj+1 ) = l!

Since X and Y are independent P(X = x and Y = y) = P(X = x) £ P(Y = y) Now P(X + Y = k) =

P(X1 + X2 + :::: + Xj = i and Xj+1 = l ¡ i)

£

25

)

P(X1 + X2 + :::: + Xj + Xj+1 = l) has PDF

i=0

a X » Po(m1 ) and Y » Po(m2 ) are two independent random variables.

m1x e¡m1

.

l!

)

l P

7 3

x!

(m1 + m2 + :::: + mj )l e¡(m1 +m2 +::::+mj )

and P(X + Y > 5) = 1 ¡ P(X + Y 6 5) ¼ 0:228 fas X + Y » Po(4:083)g 5

m1x e¡m1

b P(X 6 1) ¼ 0:9988 which is > 0:995 So, the chain is considered safe.

95

4

e¡(m1 +m2 ) £ (m1 + m2 )k k!

q p2 0:95 = (0:05)2

b Var(X) =

p ) ¾ ¼ 380 ) ¾ ¼ 19:5 throws

100

3

=

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\166IB_HL_OPT-Stat-Prob_an.cdr Tuesday, 16 April 2013 4:28:15 PM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS

e As ¹ ¼ ¾2 , n > 50 and p 6 0:1 a Poisson distribution could be used to approximate the binomial distribution where X » Po(10). i P(X 6 4) ¼ 0:0293

4 X » DU(40)

n2 ¡ 1 b Var(X) = 12 402 ¡ 1 = 12

n+1 a ¹= 2 = 41 2 = 20:5

ii P(X > 6) = 1 ¡ P(X 6 5) ¼ 0:933

= 133:25 p ) ¾ = 133:25 ¼ 11:5 5

A B C As

X » Po(6). X » Po(1). X » Po(24). B has the highest

f The Poisson approximation is reasonably good. 9

P(X = 3) ¼ 0:0892 P(X = 1) ¼ 0:3679 P(X = 17) ¼ 0:0308 probability it is the most likely to occur.

a X = a return from playing the game = ¡E14:90, ¡E14:80, ¡E14:70, ¡E14:60, ¡E14:50, ¡E14:40, ¡E14:30, E0, E15, E85

P(X = 5) =

¡4¢ 3

4

1

(0:47) (0:53)

=

¼ 0:103

3

4

3

(0:47) (0:53)

=1¡

d

£¡3¢ ¡ ¢ (0:47)4 (0:53)0 + 43 (0:47)4 (0:53)1 3 ¡5¢ ¡6¢ ¤ 4 2 4 3 (0:47) (0:53) +

3

3

(0:47) (0:53)

r = p

4 0:53

10

This is the average number of games it would take them to win without restriction, i.e., by playing as many games as they need. However, in a World Series, no more than 7 games will be played (assuming no draws) to decide the winner.

¾=

r =

¼ 1:61

b E(X) =

1¡p p2

¼ 0:994

c

P(there are empty seats) = P(X > 6) = 1 ¡ P(X 6 5) ¼ 0:918

11

magenta

i P(T = 0) ¼ 0:687 ii P(T > 2) = 1 ¡ P(T 6 2) ¼ 0:006 46

E(T ) = np = 0:375 Var(T ) = np(1 ¡ p) = 0:373 The mean and variance are almost the same which suggests that T can be approximated using a Poisson distribution. d Using T » Po(0:375) i P(T = 0) ¼ 0:687 c

yellow

25

0

ii P(T > 2) = 1 ¡ P(T 6 2) ¼ 0:006 65

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

cyan

a T = number of wrong numbers dialled in a typical week ) T » B(75, 0:005) b

ii ¾ 2 = np(1 ¡ p) = 9:20

i ¹ = np = 255 £ 0:0375 ¼ 9:56

d

7 8 1 64

= 56

8 Let X = the number who do not arrive. a Then X » (255, 0:0375) P(more than 250 arrive) = P(X 6 4) ¼ 0:0362

of success

) ¹ = 8 and ¾ ¼ 7:48 c P(X < 5) = P(X 6 4) ¼ 0:414

1 ¡ 0:62 0:622

b

1 8

1 1¡p = 8, Var(X) = p p2 =

50

1 0:62

a X » Geo( 18 ) Assumptions: ² calls are independent of each other

75

=

r

1 p

a game costing E15 the expected loss is 22 cents for a game costing E14:80 the expected loss is 2 cents the smallest amount is E14:80 . each game E(X) = ¡E1:22 for 1000 games they would expect to make 1000 £ E1:22 = E1220

² each call is made with probability

a If X is the number of attempts needed then X » Geo(0:62). This assumes that attempts are independent and the probability of getting through remains constant. b P(X > 3) = 1 ¡ P(X 6 2) ¼ 0:144 c ¹=

i For ) ) ii For )

¼ 7:547 games

95

and E(X) =

25

xi2 p ¡ [E(X)]2

c If X » DU(10), it assumes that X has values 1, 2, 3, 4, ...., 10 which is not the case here.

d X » NB(4, 0:53)

0

P

P(Redsox win) = P(X = 0, 1, 2, or 3) = 1 ¡ P(X = 4, 5, 6, or 7) +

5

1 (¡14:70) 10

= (¡14:90)2 (0:1) + (¡14:80)2 (0:1) + :::: + (85)2 (0:1) ¡ (¡0:22)2 ¼ 894

¼ 0:145

¼ 0:565

7

+

= ¡E0:22 and Var(X) =

100

c

¡6¢

pi xi 1 1 (¡14:90) + 10 (¡14:80) 10 1 + :::: + 10 (85) Euro

=

P(X = 7)

b

P

b E(X) =

6 Let X = Yankees beat Redsox in a game X » NB(4, 0:47) a

167

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\167IB_HL_OPT-Stat-Prob_an.cdr Tuesday, 16 April 2013 4:28:24 PM BRIAN

IB HL OPT 2ed Calculus

168

WORKED SOLUTIONS b As the median is 10,

e Both results are very close verifying that for large n and small p, the binomial distribution can be approximated by the Poisson distribution with the same mean i.e., X » Po(np). 12

Z

f (x) dx = 0:5 0 10

Z

¸e¡¸x dx = 0:5

)

i As 0 < q < 1,

a

10

1 1 + q + q 2 + q3 + :::: has sum to infinity 1¡q

0

)

fsum of an infinite geometric seriesg

ii Thus 1 + q + q 2 + q 3 + q 4 + :::: = (1 ¡ q)¡1 Differentiating both sides with respect to q gives: 1 + 2q + 3q 2 + 4q 3 + :::: = ¡(1 ¡ q)¡2 £ (¡1)

¡¸

e¡¸x

)

xq

x¡1

x=1

e¡10¸ = 0:5

)

¡10¸ = ln )

c

b X » Geo(p) ) P(X = x) = p(1 ¡ p)x¡1 Now E(X) = =

P 1

P

=p

P

xq

= ¡ ln 2

0:069 315e¡0:069 315x dx

0

= 0:006 80 d E(X) =

xpqx¡1

x=1 1

72

=1¡

xP(X = x)

x=1 1

1 2

ln 2 ¼ 0:0693 ¸= 10

P(seat purchased after 3 days) = P(X > 72) = 1 ¡ P(X < 72)

Z

= pq x¡1 where q = 1 ¡ p

= 0:5

0 0

)

=

1 P

i10

(¡e¡10¸ ) ¡ (¡e ) = 0:5

)

1 (1 ¡ q)2 1 = for 0 < q < 1 (1 ¡ q)2

h 1

¸

1 ¼ 14:4 hours ¸

The average time to buy a ticket is ¼ 14:4 hours. 3 X » N(¹, ¾ 2 ) If P(X > 13) = 0:4529 then

x¡1

x=1

1 (1 ¡ q)2 1 =p£ 2 p 1 = p =p£

ffrom a iig

P

³X ¡ ¹

)

´

13 ¡ ¹ = 0:5471 ¾ ¾ ³ ´ 13 ¡ ¹ P Z< = 0:5471 ¾ 13 ¡ ¹ ) = invNorm (0:5471) ¾
28) = 0:1573 then

³

EXERCISE C

´

28 ¡ ¹ = 0:8427 ¾ 28 ¡ ¹ ) ¼ 1:005 62 ¾

P Z
58) = 1 ¡ P(X 6 57) ¼ 0:994 0.3

d As np = 73:8 > 5 and n(1 ¡ p) = 106:2 > 5 we can approximate this binomial distribution to the normal distribution and X » N(73:8, 43:5) P(X > 58) = P(X ¤ > 57:5) ¼ 0:993 6

a

a X = number of mistakes made on one page X » Po(2:5) is the discrete random variable. Consider Y = X1 + X2 + X3 + X4 + :::: + X52 where Xi are assumed to be independent. E(Y ) = E(X1 ) + E(X2 ) + :::: + E(X52 ) = 52 £ 2:5 = 130 and Var(Y ) = Var(X1 ) + Var(X2 ) + :::: + Var(X52 ) = 52 £ 2:5 fassumptiong = 130 also. So, Y » Po(130) b

P(X > 2) = 1 ¡ P(X 6 2) ¼ 0:456

i

ii

3

Also (3 ¡ a)k = 0:3,

3¡a = 0:3 15

£2 ¤ k

)

x 5

1

1 , 15

4 15

0 elsewhere

9

a T » N(7, 36) ) ¾=6 P(jT ¡ 6j < 2:3) = P(¡2:3 < T ¡ 6 < 2:3) = P(3:7 < T < 8:3) ¼ 0:295 b X » B(4, 0:294 61) where X is the number of times T lies in the interval 3:7 < T < 8:3 and P(X = 2) ¼ 0:259 1 2

¡ x

10

a f (x) =

e

2

= 0:5e¡0:5x

which is the PDF of an

exponential random variable with ¸ = 0:5 i.e., X » Exp(0:5) b

i

¹X = E(X)

1 = ¸

magenta

yellow

95

ii

Var(X) =

1 ¸2

=4 ) ¾X = 2

100

50

75

25

0

5

95

100

50

75

25

0

5

95

=2

100

75

50

3¡a = 0:3 15

d CDF, F (X) =

=1

25

0

5

95

100

50

75

25

0

so

x¡a x + 1:5 = b ¡ a 15 ( 2x + 3 , ¡1:5 < x < 13:5 ) F (X) = 30

P(Y > 104) = 1 ¡ P(Y 6 104) ¼ 0:989

) 2k ¡ 2 = 5 ) k = 3:5 Thus X » U(1, 3:5).

5

1 15

=

¡1:5 6 x 6 13:5

c P(5 < X < 9) =

2k 2 ¡ =1 5 5

cyan

so

x

) b = 13:5 b f (x) =

f (x) dx = 1 )

b

0:6 9

and (3 ¡ a)k = 0:3,

k

1

12

9

Now 9k = 0:6, so k =

7 X is a uniform continuous random variable. a

0.1

) 3 ¡ a = 4:5 ) a = ¡1:5

c X » N(2:5, 2:5) ) P(X > 2) = P(X ¤ > 2:5) ¼ 0:500 Y » N(130, 130) ) P(Y > 104) = P(Y ¤ > 104:5) ¼ 0:987 The approximation for X is poor, but that for Y is very good. This is probably due to the fact that 2:5 = ¸ is not large enough.

Z

0.6

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\169IB_HL_OPT-Stat-Prob_an.cdr Wednesday, 17 April 2013 5:04:15 PM BRIAN

IB HL OPT 2ed Calculus

170

WORKED SOLUTIONS iii If the median is m, say

Z

i With u0 = e¡ax

c

m

f (x) dx = 0:5

Z )

0 m

0:5

Z0 m )

£

)

¤

m

¡2e

Z = ¡xe

m

¡ 2

e

)

¡ )

=

1 ¡ax e + constant ¡a ³ ´ 1 + constant = ¡e¡ax x + a

1 2

ii With u0 = e¡ax

m = ln 2

1 2

= ¡ ln 2

Z

m = 2 ln 2 = ln 4

a

0

Z

£

)

a

¤

)

= ¡x2 e¡ax ¡

³1´

= e¡ax

= 1:8

¡0:5a

) )

)

1 = a Z

¡¸x

F (x) = P(X 6 x) = 1 ¡ e

x

¡2

) P(X 6 x) = 1 ¡ e P(X 6 1) = 1 ¡ e¡0:5 ¼ 0:3935

11

¡1

¡0:2

= (1 ¡ e ) ¡ (1 ¡ e = e¡0:2 ¡ e¡1 ¼ 0:4509

1

E(X 2 ) =

P(0:4 6 X 6 2) = P(X 6 2) ¡ P(X 6 0:4)

ii

³

=0¡ ¡

c When X » Exp(¸), the CDF of X is

i

0

³ 1´

a = 2 ln 10 = ln 100

³

³

2 =0¡ ¡ 2 a 2 = 2 a

y

1 a

´i1 0

x2 ae¡ax dx

h0

a f (x) = ae¡ax for a = 1, 2, 3

+ constant

a

= e¡ax ¡x2 ¡

)

´

1

= ¡e¡ax x +

0:5a = ln 10

1 a

xae¡ax dx

h

e0:5a = 10

)

Z

d ¹ = E(X) =

= 0:1

e

³

£ ¡e¡ax x +

a ´ 2x 2 2 ¡x ¡ ¡ 2 + constant a a

³

¡ 1 = ¡0:9

e

v 0 = 2ax

¡2xe¡ax

= ¡x2 e¡ax + 2

e¡0:5x dx = 1:8

0 a 1 e¡0:5x ¡0:5 0 ¡0:5a

1 ¡ax e ¡a

u=

Z

0:5e¡0:5x dx = 0:9

)

v = ax2

ax2 e¡ax

)

iv We need to find a, say such that

Z

¡e¡ax dx

¡

= ¡xe¡ax +

=1

+ 2e0 = 1

)

uv0

¡ax

e¡0:5x dx = 1

¡ 2

)

axe¡ax = uv ¡

)

e¡0:5x dx = 0:5

0 m 1 e¡0:5x ¡0:5 0

v = ax

1 ¡ax u= e v0 = a ¡a Z Z

´

2x 2 ¡ 2 a a

´i1 0

and Var(X) = E(X 2 ) ¡ [E(X)]2

3

2 ¡ a2 1 = 2 a

y = 3e-3x 2

=

y = 2e-2x y = e-x

1

³ 1 ´2 a

x

Z

EXERCISE D.1

Z

1

t

ae¡ax dx = lim

b

t!1

0

= lim

0

h ³ 1 ´

t!1

1

ae¡ax dx

a

¡a

e¡ax

a X has probability distribution:

it

= lim (¡e¡at + 1)

)

=

yellow

3

1 3

1 3

95

1 (t 3

+ t2 + t3 )

t t(t3 ¡ 1) (1 + t + t2 ) or for t 2 R 3 3(t ¡ 1)

100

50

75

25

0

5

95

100

50

75

25

0

=

5

95

50

75

25

0

5

95

100

50

75

25

0

5

100

magenta

2

1 3

X » DU( 13 )

=1 f (x) is a well defined PDF.

cyan

1

and G(t) = p1 t + p2 t2 + p3 t3

0

t!1

)

x P(X = x)

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\170IB_HL_OPT-Stat-Prob_an.cdr Wednesday, 17 April 2013 10:39:37 AM BRIAN

IB HL OPT 2ed Calculus

171

WORKED SOLUTIONS b X has probability distribution: 1 t 3

) G(t) =

= 2

= 3

2 t 11

+

3 2 t 11

= 4

P(X = x)

1 3

5 7 t 11

+

b [G(t)]4 = ( 56 + 16 t)4 = = =

+

7

x

1

2

3

P(X = x)

1 5

2 5

2 5

a X has probability distribution x

1

2

3

4

5

6

P(X = x)

1 6

1 6

1 6

1 6

1 6

1 6

t (1 + 2t + 2t2 ) 5

=

x

0

1

2

3

P(X = x)

27 125

54 125

36 125

8 125

a X » B(1, p) has probability distribution:

c

i Die 2

0 1¡p

x P(X = x)

=

x

0

1

P(X = x)

1 2

1 2

d y

0

1

2

P(Y = y)

1 4

1 2

1 4

8

( 16 )4

1

2

3

4

500 1296

150 1296

20 1296

1 1296

b

yellow

4

5

6

7

8

2

3

4

5

6

7

1

2

3

4

5

6

3

4

5

6

7

8

9

10

P(U = u)

1 24

2 24

3 24

4 24

4 24

4 24

3 24

2 24

1 24

2 3 t 24

+

1 2 t 24

+

3 8 t 24

+

3 4 t 24

2 9 t 24

+

+

4 5 t 24

+

4 6 t 24

+

4 7 t 24

1 10 t 24

t2 [1 + 2t + 3t2 + 4t3 + 4t4 + 4t5 + 3t6 24

G(t)H(t)

i P(X = 1) = p = p1 =

1 4

ii P(X = 2) = p(1 ¡ p) =

50

iii P(X = k) =

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

3

1

a X is the number of independent trials needed to get a successful outcome. Each trial has probability 14 of being successful ) X » Geo( 14 )

625 0 500 1 150 2 20 3 1 t + 1296 t + 1296 t + 1296 t + 1296 t4 1296 1 2 3 4 [625 + 500t + 150t + 20t + t ] 1296

25

2

fusing synthetic multiplicationg

0

0

9

= K(t)

625 1296

5

95

10

8

t t (1 + t + t2 + t3 ) £ (1 + t + t2 + t3 + t4 + t5 ) 6 4 t2 (1 + t + t2 + t3 )(1 + t + t2 + t3 + t4 + t5 ) = 24 t2 = [1 + 2t + 3t2 + 4t3 + 4t4 + 4t5 + 3t6 + 2t7 + t8 ] 24

x

100

50

75

25

0

5

9

7

=

P(X = x)

magenta

8

6

+ 2t7 + t8 ]

1 0 t + 12 t1 2 1 (1 + t) 2

which is

cyan

7

5

2

=

X » B(4, 16 ) and X has probability distribution ( 56 + 16 )4 = ( 56 )4 + 4( 56 )3 ( 16 ) + 6( 56 )2 ( 16 )2 + 4( 56 )( 16 )3 +

) H(t) =

6

4

+

1 0 t + 12 t1 + 14 t2 4 1 = 4 (1 + 2t + t2 ) 1 (1 + t)2 = [G(t)]2 4

) H(t) =

5

3

U

ii K(t) =

3 + 2t 5

95

a

4

1 3 k¡1 ( ) 4 4

1 4

£

100

6

t (1 + t + t2 + t3 ) 4

U = X + Y has probability distribution:

) H(t) = c H(t) =

t (1 + t + t2 + t3 + t4 + t5 ) 6

Die 1

1 p

b For X » B(1, 0:4), G(t) = 0:6 + 0:4t

b Y has probability distribution:

+ 16 t2 + 16 t3 + 16 t4 + 16 t5 + 16 t6

H(t) =

27 54 36 2 8 3 + 125 t + 125 t + 125 t 125 3 2 1 (8t + 36t + 54t + 27) 125 1 (2t + 3)3 125

=

1 t 6

b Similarly for Y ,

2 ) 5

) G(t) =

20t3 + t4 )

= H(t)

) G(t) =

a X has probability distribution:

+ 4(5)t3 + t4 )

1 12 t 11

) G(t) = p0 + p1 t = (1 ¡ p) + pt = 1 ¡ p + pt

5

1 (5 + t)4 64 1 (54 + 4(5)3 t + 6(5)2 t2 1296 1 (625 + 500t + 150t2 + 1296

+ 25 t2 + 25 t3

1 t 5

X has probability distribution:

=

5

1 3

t (2 + 3t + 5t6 + t11 ), t 2 R 11

a X » B(3,

b G(t) =

2

1 3

+ 13 t2 + 13 t5

a X has probability distribution: b G(t) =

1

t (1 + t + t4 ), t 2 R 3

= c G(t) =

x

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\171IB_HL_OPT-Stat-Prob_an.cdr Tuesday, 16 April 2013 5:31:55 PM BRIAN

3 4

IB HL OPT 2ed Calculus

172

WORKED SOLUTIONS c G(t) = = = =

P

pi ti k=1 1 t + 14 ( 34 )t2 4 1 t 1 + ( 34 t) 4

¡

µ

1 t 4

+ ( 14 )( 34 )2 t3 + ::::

for jtj
10 we can ³ ´ 0:04 £ 0:96 assume that p b » N 0:04, 1000 P(p b > 0:07) ¼ 6:46 £ 10¡7 b With such a small probability we would reject the egg producers claim.

X » N(18:2, 3:32 )

¾ 3:3 = p n 25

a ¹X = 18:2, ¾ X = p

i Claim is: each bar averages 17 g ii Since X » N(18:2,

c

P(X < 17) ¼ 0:0345 3:45% fail to meet the claim.

i ¹X = 18:2 and ¾ X = ii Since X » N(18:2, )

17

a

By the CLT S » N(315,

0

4 ) 15

25 ) 10

b< and P(p

p5 10

3

ii ¾ =

b p

p = np(1 ¡ p) p

100 £ = ¼ 4:3301

4:33012 ) 100

0:465 £ 0:535 2500

d

0.000225

3 4

£

4

³

p b » N 0:85,

1 4

c

´

yellow

)

50

d

25

0

i p b » N 0:85,

0:85 £ 0:15 200

´

P(p b < 0:75) ¼ 0:000 037 4

ii P(0:75 < p b < 0:87) ¼ 0:786

5

95

0:85 £ 0:15 n

b For p b to be approximated by the normal distribution we require that np > 10 and n(1 ¡ p) > 10 ) 0:85n > 10 and 0:15n > 10 ) n > 11:76 and n > 66:67 ) n > 67

100

50

75

m + 3¾ ¼ 0.495 < 0.50

p b§ 3:5% lies outside the range ¹ § 3¾, so this probability is extremely small. a p = 0:85 and for n sufficiently large

³

25

0

5

95

magenta

0.000225 0.465

fCLTg

100

50

r

b < 0:43 or pb > 0:50) = P(p = 1 ¡ P(0:43 6 p b 6 0:50) ¼ 0:000 451

75

¾X

75

25

0

5

95

100

50

75

25

0

p(1 ¡ p) = n

P(p b differs by more than 3:5% from p)

iii

) P(X 6 0:68) ¼ 0:0530 b From a iv, the probability of getting 68 cured patients in a sample of 100 is very low (about 5:3%). This suggests that either the sample was biased or the company’s claim of a 75% cure rate is not justified.

5

1 ) 490

i P(p b < 0:46) ¼ 0:308 ii P(0:45 < p < 0:47) ¼ 0:626

c

iii P(X 6 68) ¼ 0:0693

cyan

A

b p b » B(2500, 0:465) and is approximated by N(0:465, 0:009 982 )

i Let X = the number of people cured then X » B(100, 34 ).

iv X » N(0:75,

100

1

¼ 0:009 98

and Var(U ) = Var(E) + 9E(S) 4 = 25 + 9( 15 ) 10 = 4:9 U » N(5, 4:9) and P(U < 0) ¼ 0:0119 c Let the contents of the three small cartons be S1 , S2 , and S3 and consider V = E ¡ (S1 + S2 + S3 ) E(V ) = E(E) ¡ E(S1 ) ¡ E(S2 ) ¡ E(S3 ) = 950 ¡ 315 ¡ 315 ¡ 315 =5 Var(V ) = Var(E) + Var(S1 ) + Var(S2 ) + Var(S3 ) = 25 + 3(4) = 37 ) V » N(5, 37) P(E < S1 + S2 + S3 ) = P(E ¡ (S1 + S2 + S3 ) < 0) = P(V < 0) ¼ 0:206

3 4

£ 57

¼ 0:538

b p r

P(E < 3S)

ii ¹X = np = 100 £ = 75

29 ) 100

2 7

i p = 0:465 and n = 2500 ) ¹ = p = 0:465

a

= P(E ¡ 3S < 0)

a

p b » N( 27 ,

)

Let U = E ¡ 3S ) E(U) = E(E) ¡ 3E(S) = 950 ¡ 3(315) =5

18

p b » N @ 27 ,

assume that

p2 15

ii E has ¹E = 950 and ¾ E = b

n = 100

As np ¼ 28:57 and n(1 ¡ p) ¼ 71:43 are both > 10 we can

P(X < 425 ) ¼ 0:002 09 26 about 0:209% fail to meet the claim now.

E » N(950,

2 , 7

3:32 ), 26

i S has ¹S = 315 and ¾ S =

)

2 p=

3:3 p 26

i n = 500, p b = 350 = 0:7 and p = 0:85 500 np = 425 and n(1 ¡ p) = 75 are both > 10

95

)

3:32 ), 25

100

b

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\178IB_HL_OPT-Stat-Prob_an.cdr Thursday, 18 April 2013 11:42:11 AM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS

³

Now p b » N 0:85, )

P(p b6

350 ) 500

0:85 £ 0:15 500

¼0

´

2

5 n = 400, p b » N @ 25 ,

2 5

a x10 has distribution

X1 + X2 + X3 + :::: + X10 with 10 ¾2 E(X 10 ) = ¹ and Var(X 10 ) = 10

X 10 =

fusing technologyg

ii Under the given conditions there is virtually no chance of this happening. This means that either: (1) it was a freak occurrence which is possible but extremely unlikely (2) the population proportion was no longer 85% (probably < 85%) (3) the sample was not taken from the area mentioned.

0

£ 35

400

Likewise x25 has distribution

X1 + X2 + :::: + X25 with E(X 25 ) = ¹ and 25 2 ¾ Var(X 25 ) = and so Var(X 25 ) < Var(X 10 ) 25

X 25 =

) X25 is a more efficient estimator of ¹ and so is preferred. b The larger the sample size n, the smaller the value of

1 A

Var(X) =

6

150 ) 400

b P(p b
10

³

b » N 0:9, ) p

0:9 £ 0:1 250

´

P(p b6

=

) E(T1 ) = = )

T2 = ) E(T2 ) = = )

Var(T2 ) = =

yellow

y = f(¸)

1 2

)

f (¸) is least when ¸ =

)

Var(T ) is least when ¸ =

)

the most efficient estimator of T has ¸ =

1 2

X1 + X2 T = 2

1 2

and is

95

100

50

75

25

0

= the sample mean

5

95

100

50

75

25

0

5

95

100

50

75

ai = 1

= (¸2 + 1 ¡ 2¸ + ¸2 )¾ 2 = (2¸2 ¡ 2¸ + 1)¾ 2 To get the most efficient estimator of T of this form we need to minimise Var(T ) and hence 2¸2 ¡ 2¸ + 1 (as ¾ 2 is constant). Consider f (¸) = 2¸2 ¡ 2¸ + 1 f 0 (¸) = 4¸ ¡ 2 ) f 0 (¸) = 0 , ¸ = 12

2 X + 16 X2 + 36 X3 6 1 2 E(X1 ) + 16 E(X2 ) + 36 E(X3 ) 6 2 ¹ + 16 ¹ + 36 ¹ 6

25

0

n P

b Var(T ) = ¸2 Var(X1 ) + (1 ¡ ¸)2 Var(X2 ) = ¸2 ¾ 2 + (1 ¡ ¸)2 ¾ 2

16 2 9 25 2 ¾ + 144 ¾ 2 + 144 ¾ 144 50 2 ¾ 144 ( 26 )2 ¾ 2 + ( 16 )2 ¾ 2 + ( 36 )2 ¾ 2 14 2 ¾ 36 56 2 ¾ ) 144

5

95

100

50

75

25

0

5

ai

4 If T = ¸X1 + (1 ¡ ¸)X2 , 0 6 ¸ 6 1 a E(T ) = ¸E(X1 ) + (1 ¡ ¸)E(X2 ) = ¸¹ + (1 ¡ ¸)¹ =¹ ) T is an unbiased estimator for ¹.

4 3 5 X + 12 X2 + 12 X3 12 1 4 3 5 E(X1 ) + 12 E(X2 ) + 12 E(X3 ) 12 4 3 5 ¹ + ¹ + ¹ 12 12 12

magenta

P

i=1

(= ) Var(T1 ) < Var(T2 ) ) T1 is a more efficient estimator of ¹ than T2 .

cyan

ai µ

and E(T ) = µ ,

4 2 3 2 5 2 c Var(T1 ) = ( 12 ) Var(X1 ) + ( 12 ) Var(X2 ) + ( 12 ) Var(X3 )

=

P

ai E(Ti )

i=1

=¹ T2 is an unbiased estimator of ¹. =

i=1 n

=µ

which is almost zero

=¹ T1 is an unbiased estimator of ¹.

b

n P

i=1 n

EXERCISE F T1 =

ai Ti

) E(T ) =

b 6 0:8 is virtually impossible. ) the chance that p c As the chance is practially zero there is little chance that the manufacturers claim is correct.

a

n P

i=1

200 ) 250

b 6 0:8) = P(p ¼ 6:82 £ 10¡8

1

E(T1 ) = E(T2 ) = µ E(T ) = aE(T1 ) + bE(T2 ) = aµ + bµ = (a + b)µ and E(T ) = µ , a + b = 1

b T =

) p b » N(0:9, 0:000 36) Assumptions made are: (1) the approximation to normal is satisfactory (2) When selected at random, the life of any tyre is independent of the life of any other tyre. b

¾2 . Thus, as n increases, the more efficient n

the estimator X is of ¹.

) p b » N(0:4, 0:0006) a P(p b>

179

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\179IB_HL_OPT-Stat-Prob_an.cdr Wednesday, 17 April 2013 4:09:23 PM BRIAN

IB HL OPT 2ed Calculus

180 5

WORKED SOLUTIONS a E(T ) = E( 39 SX2 + 69 SY2 ) = = =

= E(SX2 ) + E(SY2 ) = Var(X) + Var(Y ) = Var(X + Y ) = Var(U ) ) sX2 + sY2 is an unbiased estimate of Var(U).

=¾ b Let t =

(n ¡ 1)sX2 + (m ¡ 1)sY2

b U = aX + bY , a, b 2 R + i E(X) = E(aX + bY ) = aE(X) + bE(Y ) = a¹X + b¹Y Var(U ) = Var(aX + bY ) = a2 Var(X) + b2 Var(Y ) = a2 ¾ X2 + b2 ¾ Y2

n+m¡2 1 ) E(T ) = [(n ¡ 1)E(SX2 ) + (m ¡ 1)E(SY2 )] n+m¡2 1 [(n ¡ 1)¾ 2 + (m ¡ 1)¾ 2 ] = n+m¡2 fas SX2 and SY2 are unbiased estimates of ¾ 2 g

1 ¾ 2 (n + m ¡ 2) n+m¡2

=

ii E(aX + bY ) = aE(X) + bE(Y ) = a¹X + b¹Y = E(U )

= ¾2 6

E(SX2 + SY2 )

iii

3 E(SX2 ) + 69 E(SY2 ) 9 3 2 2 E(S4¡1 ) + 69 E(S7¡1 ) 9 3 2 6 2 ¾ + 9¾ 9 2

a X » N(¹, ¾ 2 )

aX + bY is an unbiased estimator for E(U ) ax + by is an unbiased estimate for E(U )

) )

X1 + X2 + X3 + :::: + Xn X= n 1 1 1 ) E(X) = E(X1 ) + E(X2 ) + :::: + E(Xn ) n n n 1 = (¹ + ¹ + :::: + ¹) fn ¹sg n 1 = n¹ n

E(aSX2 + bSY2 )

iii

= aE(SX2 ) + bE(SY2 ) = aVar(X) + bVar(Y ) 6= Var(U ) ffrom b ig ) asX2 + bsY2 is generally not an unbiased estimate of Var(U).

=¹

1 1 1 Var(X1 ) + 2 Var(X2 ) + :::: + 2 Var(Xn ) n2 n n 1 = 2 (¾ 2 + ¾ 2 + :::: + ¾ 2 ) n n¾ 2 = 2 n ¾2 = n

x = 4:39 SX ¼ 2:793 58 = sn¡1 ¾ X ¼ 2:550 18 = sn a An unbiased estimate of ¹ is x = 4:39

8 For the sample

Var(X) =

b An unbiased estimate of ¾ 2 is ¼ 2:793 582 ¼ 7:8041 c sn2 ¼ (2:550 18)2 ¼ 6:5034

But Var(X) = E(X ) ¡ [E(X)]2

¾2 2 = E(X ) ¡ ¹2 n ¾2 ¾2 2 ) E(X ) = ¹2 + where >0 n n

n sn2 = 65 sn2 n¡1 sn2¡1

2 d sn¡1 =

2

Check:

)

sn2

9 X » U(0, b)

Qx

¼ 1:2 X

f(x)

2

) E(X ) > ¹2 2

b As E(X ) 6= ¹2 , X 7 E(X) = ¹X , E(Y ) = ¹Y , a

2

is a biased estimator of ¹2 .

Z

Var(X) = ¾X2 Var(Y ) =

a E(X) =

¾Y2

E(X + Y )

¸b

1 x2 b 2

b X is a sample mean estimator of Now E(2X) = 2E(X) = 2

yellow

0

95

b 2

³b´ 2

=b

2X is an unbiased estimator of b for all n 2 Z + .

100

50

75

25

0

5

95

)

100

50

75

25

0

5

95

50

75

25

0

5

95

100

50

75

25

0

5

100

magenta

0

1 x dx b

b = 2

= E(X) + E(Y ) = ¹X + ¹Y = E(U ) ) x + y is an unbiased estimate of E(U ).

cyan

· =

= ¾ X2 + ¾ Y2 ii

b

x f (x) dx = 0

i U =X +Y E(U) = E(X) + E(Y ) = ¹X + ¹Y and Var(U) = Var(X) + Var(Y )

x

b

Z

b

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\180IB_HL_OPT-Stat-Prob_an.cdr Thursday, 18 April 2013 11:43:31 AM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS 10 S¹2 = = = =

1 n 1 n 1 n 1 n

1 = n =

·

n P

¸

i=1

·

n P n P

¸

(Xi2 ¡ 2¹Xi + ¹2 ) Xi2 ¡ 2¹

i=1

·

n P

n P

P n

µ

Xi + n¹2

¸

=

Xi2 ¡ 2¹(n¹) + n¹2

¸ 2

Xi ¡ n¹

i=1 n

µ

) E(S¹2 ) = =

n P 1 E Xi2 n i=1

1 n

1 = n

·

n P n P

¶ ¡ ¹2

E(Xi2 ) ¡ ¹2

¸

=

1 mn

(Var(Xi ) + [E(Xi )] ) ¡ ¹2

i=1

i=1

E(SA2 ) =

E(SB2 ) =

b) = E(pb2 ) ¡ [E(pb)]2 But Var(p µ

97 30

) E

2 sn¡1 =

4 2 s 3 n 2 2 ) sn = 34 sn¡1 3 3 2 ) = ¾2 E(Sn¡1 4 4

8 2 ) E(Sn¡1 9

=

¶

p bqb n

= =

30

8 2 ¾ 9

µ

c Since E

(n1 ¡ 1) + (n2 ¡ 1) + (n3 ¡ 1) + :::: + (nr ¡ 1)

¡r

n

¾

2

b p

6= ¾ 2 ,

b p

n n¡1

³n ¡ 1´ n

¾2

b p

yellow

95

50

75

25

0

5

95

100

50

75

25

0

5

95

100

magenta

p bqb is an unbiased estimate of ¾ 2 . b p n¡1

100

)

75

50

¸

b p

¶

25

0

5

95

100

50

75

25

0

¶

p bqb n

³n ¡ 1´

=¾2

i=1

5

·

=

ni si2

ni

1 p(1 ¡ p) p¡ ¡ p2 n n

p bqb is a biased estimate of ¾ 2 . b p n ³ n ´ pbqb p bqb d As = , n¡1 n µ ¶ n¡1 µ ¶ p n p bqb bqb E = E n¡1 n¡1 n

¾2

n1 s12 + n2 s22 + n3 s32 + :::: + nr sr2

i=1 t= µ n P

=

=

= ¾2 t is an unbiased estimate of ¾ 2 .

r P

¤ 1£ E(p b) ¡ fVar(pb) + [E(pb)]2 g n

np ¡ p + p2 ¡ np2 n2 (n ¡ 1)p ¡ (n ¡ 1)p2 = 2 µ n ¶ n ¡ 1 p(1 ¡ p) = n n

19 2 ) = 19 ¾ 2 E(Sn¡1 20 20 4 2 9 E(SA ) + 30 E(SB2 ) + 20 E(SC2 ) 30 30 4 3 2 9 8 2 ( ¾ ) + 30 ( 9 ¾ ) + 20 ( 19 ¾ 2 ) 30 4 30 20

³ 3 + 8 + 19 ´

=

=

E(SC2 ) = ) E(T ) =

¶

b) ¡ E(pb2 ) = E(p

For sample C,

cyan

p(1 ¡ p) n

b E(p bb q ) = E(p b(1 ¡ pb)) = E(p b ¡ pb2 )

Likewise for sample B,

or

j=1

= ¾2

2 fso sn¡1 is an unbiased estimate of ¾ 2 g

b t=

¶ E(Yj )

a E(p b) = p and E(b q) = E(1 ¡ p b) = 1 ¡ E(p b) =1¡p

ii For sample A,

)

¶µ m P

E(Xi )

1 = (n¾ 2 + n¹2 ) ¡ ¹2 n

4(3) + 9(5) + 20(2) i t= = 30

)

n P

13 For n sufficiently large p b » N p,

S¹2 is an unbiased estimator of ¾ 2 .

a

µ

fas Xi and Yj are independentg

µ

2

= ¾2 + ¹2 ¡ ¹2

11

n P m 1 P E(Xi )E(Yj ) mn i=1 j=1

= ¹X ¹Y ) x y is an unbiased estimate of ¹X ¹Y . Note: In order to see this argument clearly work it through with x = 2, m = 3 say.

¸

fVar(X) = E(X 2 ) ¡ [E(X)]2 g

)

n P m 1 P (E(Xi Yj )) mn i=1 j=1

1 = (n¹x )(m¹Y ) mn

i=1

·

¸

¶

=

2

1 P 2 X ¡ ¹2 n i=1 i

n m 1 P 1 P Xi £ Yj n i=1 m j=1

n m P P 1 = E Xi £ Yj mn i=1 j=1

¸

i=1

i=1

·

·

=E

i=1

·

E(X Y )

12

(Xi ¡ ¹)2

181

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\181IB_HL_OPT-Stat-Prob_an.cdr Thursday, 18 April 2013 11:45:26 AM BRIAN

IB HL OPT 2ed Calculus

182

WORKED SOLUTIONS b 98% confidence interval for ¹ is

EXERCISE G.1

¾ ¾ < ¹ < x + 2:326 p n n

1 ¹ is unknown, ¾ = 10, n = 35, x = 28:9 a Using a Z-distribution, the 95% confidence interval is

x ¡ 2:326 p

Using technology this is 8:33 < ¹ < 9:08

¾ ¾ < ¹ < x + 1:960 p n n

x ¡ 1:960 p

7 n = 75, x = 513:8, sn = 14:9

r

Using technology, 25:6 < ¹ < 32:2 b Using a Z-distribution, the 99% confidence interval is

sn¡1 =

¾ ¾ < ¹ < x + 2:576 p n n

x ¡ 2:576 p

) sn¡1 ¼ 15:00 funbiased estimate of ¾g As we had to estimate ¾ using sn from the sample, the t-distribution applies. The 99% confidence interval is 509:2 < ¹ < 518:4

Using technology, 24:5 < ¹ < 33:3 c The confidence interval width becomes larger. 2 When increasing the level of accuracy we increase the interval width (as shown in question 1). We can estimate ¹ in a narrower interval but with less certainty.

8 n = 42, x = 38:2, sn = 4:7

r

3 Sample size n, ¾ = 11, x = 81:6 a

p n 75 sn = £ 14:9 74 n¡1

sn¡1 =

i n = 36, Z-distribution

p n 42 sn = £ 4:7 41 n¡1

) sn¡1 ¼ 4:757 funbiased estimate of ¾g As we had to estimate ¾ from sn , the t-distribution applies. The 90% confidence interval is 37:0 < ¹ < 39:4

¾ ¾ < ¹ < x + 1:960 p n n

x ¡ 1:960 p

Using technology, 78:0 < ¹ < 85:2 ii n = 100 Using technology, 79:4 < ¹ < 83:8 b As n increases, the width of the confidence interval decreases.

9 n = 60, x = 84:6, sn = 16:8

r

sn¡1 =

4 Using the Z-distribution, the 95% confidence interval for ¹ is

¾ ¾ < ¹ < x + 1:960 p where a = 1:960. n n

p n 60 sn = £ 16:8 59 n¡1

) sn¡1 ¼ 16:94 funbiased estimate of ¾g As we had to estimate ¾ from sn , the t-distribution applies.

x ¡ 1:960 p

a If P = 99, we need to solve P(Z < a) = 0:995

i 95% confidence interval is 80:2 < ¹ < 89:0 ii 99% confidence interval is 78:8 < ¹ < 90:4 b As n = 50, n is sufficiently large to use the normal confidence interval a

99%

5 n = 50, standard deviation = ¾, x = 38:7, Z-distribution 95%

¾ ¾ < ¹ < 84:6 + 1:960 p n n ¾ ¾ ) ¡1:960 p < ¹ ¡ 84:6 < 1:960 p n n ¾ ) j¹ ¡ 84:6j < 1:960 p n 16:94 Thus 1:960 £ p

5

¾ ¾ < ¹ < x+1:960 p n n

) n > 44:1 ) a sample of 45 or more is needed.

0.5%

0.5% 0

-a

) 84:6 ¡ 1:960 p

Z

a

Using technology gives a ¼ 2:576 b If P = 80, we need to solve P(Z < a) = 0:9 Using technology gives a ¼ 1:282 c If P = 85, we need to solve P(Z < a) = 0:925 ) a ¼ 1:440 d If P = 96, we need to solve P(Z < a) = 0:98 ) a ¼ 2:054 confidence interval for ¹ is x¡1:960 p

a For ¾ = 6, using technology, 37:0 < ¹ < 40:4 b For ¾ = 15, 34:5 < ¹ < 42:9 c As ¾ increases, the width increases.

10

P

P

a x=

6 n = 167, x = 8:7, 2:6 6 X 6 15:1 a range = 15:1 ¡ 2:6 = 12:5 and ¾ ¼ range ¥ 6 ¼ 2:083

=

95

50

25

0

5

95

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

100

yellow

P

b sn2 = =

x2 ¡ x2 n

1325:31 10

= 5:9685

100

Z

3

x2 = 1325:31

¡ 11:252

r

n sn n¡1

gives sn¡1 ¼ 2:575 and sn¡1 is an unbiased estimate of ¾. c As sn¡1 is used as an unbiased estimate of ¾ 2 the t-distribution applies. From technology, 9:96 < ¹ < 12:7

75

0

magenta

n

112:5 10

P

sn¡1 =

So, nearly all scores lie in the interval [¡3, 3] which has length 6 standard deviations.

cyan

x

= 11:25

99.73% lie between -3 and 3 for the Z distribution

-3

x = 112:5 and

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IB HL OPT 2ed Calculus

WORKED SOLUTIONS 11 Using a Z-distribution with ¾ = 17:8, the 98% confidence

b ¾ is unknown so we use the T -statistic i 95% confidence interval for ¹ is: ¡0:0371 6 ¹ 6 0:5621 ii 99% confidence interval for ¹ is: ¡0:181 6 ¹ 6 0:706 c Both confidence intervals in b contain the value 0 and also negative values. So, it is possible that ¹ < 0 at both 95% and 99% levels of confidence. That is, there is insufficient evidence at these levels to support the manufacturers claim. d At a level of confidence ®, the confidence interval is:

¾ ¾ interval for ¹ is x ¡ 2:326 p < ¹ < x + 2:326 p n n ¾ ) j¹ ¡ xj < 2:326 p n 2:326 £ 17:8 Hence

3

)

a n = 48, sn2 = 22:09 sn¡1 = =

p

n sn n¡1 48 47

£ 22:09

= 4:7497:::: ¼ 4:75 ) sn¡1 = 4:75 is an unbiased estimate of ¾. b As n is large the 99% confidence interval can be obtained using the Z-distribution. The 99% confidence interval for ¹ is

1

a b c d

2

EXERCISE G.2

C 7

D 5

E 3

F 2

G ¡1

H 11

I 6

J 5

K 4

a n = 11, u = 4:182 and sn¡1 = 3:219 ¾ is unknown, so we use the T -statistic

U ¡¹

T =

A

B

C

D

E

F

G

H

d

0:2

0:6

¡0:2

0:8

0:2

¡0:2

0:5

0:2

yellow

a A Type I error would result if X and Y are determined to have different effectiveness, when in fact they have the same. b A Type II error would result if X and Y are determined to have the same effectiveness, when in fact they have different effectiveness.

5

a H0 : H1 : b H0 : H1 :

new new new new

globe globe globe globe

has has has has

mean mean mean mean

life life life life

80 hours > 80 hours 80 hours < 80 hours

¾ ¾ 6 ¹ 6 x + 1:96 p n n ¾ ¾ x ¡ 1:96 p 6 ¹ and ¹ 6 x + 1:96 p n n ¾ ¾ x 6 ¹ + 1:96 p and x > ¹ ¡ 1:96 p n n ¾ ¾ ¹ ¡ 1:96 p 6 x 6 ¹ + 1:96 p n n x ¡ 1:96 p

1 ) )

2

50

75

25

0

5

95

100

50

ii a Type I error

a For ® = 0:05, z® ¼ 1:645 and z ® ¼ 1:960 2

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

ii a Type II error

4

2 a n = 8, d = 0:2625, sn¡1 ¼ 0:128 393

magenta

i a Type I error i a Type II error

)

Pair

cyan

a b

EXERCISE H.2

i 95% confidence interval for ¹ is 2:02 6 ¹ 6 6:34 ii 90% confidence interval for ¹ is 2:42 6 ¹ 6 5:94 b The sample of 11 is extremely small and the mean improvement u < 5 km h¡1 . There is insufficient evidence to accept the sports commission claim as values < 5 lies within both the 95% and 90% confidence intervals for ¹. 2

A Type I error involves rejecting a true null hypothesis. A Type II error involves accepting a false null hypothesis. The null hypothesis is a statement of no difference. The alternative hypothesis is a statement that there is a difference.

6 H0 : new design has top speed of 26:3 knots H1 : new design has top speed > 26:3 knots

is T » t(10)

sn¡1 p 11

2

® ¼ 0:026 513 2

a The alternative hypothesis (H1 ) would be that the person on trial is guilty. b a Type I error c a Type II error

and consider U = X2 ¡ X1 . B 1

t ® ¼ 2:3247

0:128 p 393 8

3

1 Let X1 = the throwing distance of a 13 year old X2 = the throwing distance of the same 12 year old A 3

)

p

2

EXERCISE H.1

¾ ¾ < ¹ < x + 2:576 p n n ¾ ) j¹ ¡ xj < 2:576 p n 2:576 £ 4:7497 < 1:8 p n p 2:576 £ 4:7497 n> ) 1:8 ) n > 46:20 ) n should be at least 47.

Age u

t® =

) ® ¼ 0:0530 (5:3%) we have a 94:7% confidence level.

)

x ¡ 2:576 p

) we require

)

)

95

r

100

12

sn¡1 sn¡1 6 ¹ 6 d + t® p n n 2 2 sn¡1 d + t® p ¼ 0:557 n 2 0:557 ¡ 0:2625

d ¡ t® p

) n > 190:46:::: should sample at least 191 packets.

)

183

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IB HL OPT 2ed Calculus

184

WORKED SOLUTIONS 6 (1) H0 : ¹ = 100 and H1 : ¹ < 100 (2) As ¾ is known (¾ = 1:6 g) the null distribution is Z.

b For ® = 0:01, z® ¼ 2:326 and z ® ¼ 2:576 2

3

a

¾2

(3) The test statistic is z ¤ =

= 15:79, n = 36, x = 23:75

x ¡ ¹0

¤

z =

i

x ¡ ¹0 p¾

n

p¾ n

=

99:4 ¡ 100 1:6 p

23:75 ¡ 25

¤

) z =

40

p

¼ ¡2:372 (4) We reject H0 if z ¤ lies in the critical region. (5)

15:79 6

) z ¤ ¼ ¡1:887

ii The null distribution is Z » N(0, 1) iii The p-value = P(Z 6 ¡1:887) ¼ 0:0296 fusing technologyg b

0.05

i

Z

-1.645

z ¤ = -2.372

Z

¡z® = -1.645

z¤

(6) Since z ¤ lies in the critical region we reject H0 in favour of H1 . (7) We have sufficient evidence to accept H1 , that the mean weight is less than 100 g net. There is evidence that the machine which delivers the nuts needs to be adjusted to allow more nuts into each bag.

As z ¤ < ¡z® we reject H0 in favour of H1 . ii As the p-value < 0:05 we reject H0 in favour of H1 . 4

a

i z > z® , so z > 1:645 ii z > z® , so z > 2:326

b

i z < ¡z® , so z < ¡1:645 ii z < ¡z® , so z < ¡2:326

c

i

(3) The test statistic is z ¤ =

® = 0:025 2

=

z < ¡z ®

or

) z < ¡1:960

or

z > 1:960

z < ¡z ®

or

z > z®

) z < ¡2:576

or

z > 2:576

z > z®

2

p¾

=

83:1 ¡ 80 12:9 p

¼ 3:398

200

8 ¾ 2 = 2:25 is known, n = 8, x = 1001 (1) H0 : ¹ = 1000 g and H1 : ¹ > 1000 g (2) As ¾ 2 is known we use a Z-distribution.

d Reject H0 if either z ¤ lies in the critical region of Z or the p-value is < 0:01. e

(3) The test statistic is z ¤ = 0.01

=

Z 3.398 = z ¤

yellow

p 2:25 p

¼ 1:886

95

or

100

50

(4) We reject H0 if: z ¤ lies in the critical region

25

0

1001 ¡ 1000 8

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

p¾

p 1£ 8 = p 2:25

f As z ¤ lies within the critical region we reject H0 in favour of H1 . or As the p-value = P(Z > z ¤ ) = 0:000 339 is < 0:01 we reject H0 . g We accept that ¹ > 80 at the 1% level of significance. P(Type I error) = 0:01.

magenta

x ¡ ¹0 n

75

2.326

5

2:89 p

80

2

n

cyan

21:2 ¡ 22:3

¼ ¡3:404 (4) We reject H0 if the p-value < 0:05 (5) p-value = P(Z < ¡3:404 or Z > 3:404) ¼ 0:000 663 (6) Since p-value < 0:05 we reject H0 in favour of H1 . (7) There is sufficient evidence at a 5% level to suggest that the mean fleece diameter differs in 2012 from 2008. b With x = 21:2, ¾ = 2:89, n = 80 The 95% confidence interval is 20:57 < ¹ < 21:83 and ¹0 = 21:2 lies within it. This confirms, at a 95% level of confidence that there is a significant difference in the means between 2008 and 2012.

5 ¾ = 12:9, n = 200, x = 83:1 a H0 : ¹ = 80 and H1 : ¹ > 80 b The null distribution is Z with ¾ = 12:9

x ¡ ¹0

p¾

2

® = 0:005 2

c z¤ =

x¡¹ n

2

ii

a (1) H0 : ¹ = 22:3 (no difference) H1 : ¹ 6= 22:3 (a difference) (2) Assuming ¾ = 2:89 is constant the null distribution is Z.

7

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the p-value < 0:01

IB HL OPT 2ed Calculus

WORKED SOLUTIONS (5)

or

d p-value = P(T < ¡11:82) = 0 which is < 0:02 e We reject H0 that ¹ = $13:45 at a 2% level. That is, we accept the claim that the mean price has fallen. Note: P(Type I error) = 0:02 .

p-value = P(Z > 1:886) ¼ 0:029 67

0.01

4 x = 499 mL and sn = 1:2 mL As ¾ is unknown, we use sn¡1 as an unbiased estimate of ¾

Z z® = 2.326 z ¤ = 1.886

r

(6) As z ¤ is not in the critical or As p-value is not region we do not reject < 0:01 we do not H0 . reject H0 . (7) We conclude that there is insufficient evidence to support the overfilling claim at a 1% level. (However we may be making a Type II error.)

and sn¡1 = =

2

2 H0 : ¹ = 18:5 and H1 : ¹ 6= 18:5 a i n = 24, x = 17:14, sn = 4:365

p

n sn n¡1 24 23

£ 4:365

-2.947

¼ 4:459 which is an unbiased estimate of ¾ and t¤ =

x ¡ ¹0 sn¡1 p n

¼

4:459 p 24

ii The null distribution is t(23). iii The p-value = 2 £ P(T > jt¤ j) = 2 £ P(T > 1:494) ¼ 0:1487 b

£ 1:2

x ¡ ¹0 sn¡1 p n

=

5

i

a n = 60, x = 242:6 mg, sn = 7:3 mg As ¾ is unknown, we use sn¡1 as an unbiased estimate of ¾ and sn¡1 = =

T

t = -1.494

2

=

p

0.025

£ $0:25 -2.001

sn¡1 p n

¼

242:6 ¡ 250 7:3616 p 60

or

the p-value < 0:05

or

p-value = 2 £ P(T > jt¤ j) = 2 £ P(T > 7:7864) ¼ 1:26 £ 10¡10

0.025 2.001

T

magenta

yellow

95

50

75

25

0

5

95

100

50

75

25

0

5

95

t¤

100

50

75

25

0

5

95

100

50

75

25

0

5

x ¡ ¹0

(5)

¼ $0:2503 c t¤ ¼ ¡11:82

cyan

£ 7:3

) t¤ ¼ ¡7:7864 (4) We reject H0 if: t¤ lies in the critical region

n sn n¡1 389 388

60 59

100

sn¡1 =

t¤ =

(3)

a H0 : ¹ = $13:45, H1 : ¹ < $13:45 b As ¾ is unknown we use a t-distribution sn¡1 is used as an unbiased estimate of ¾ and

r

p

n sn n¡1

¼ 7:3616 (1) H0 : ¹ = 250 mg and H1 : ¹ 6= 250 mg (2) As ¾ is unknown we use the t-statistic t(59).

t ® = 2.602

¤

As t¤ does not lie in the critical region we do not reject H0 as there is insufficient evidence to do so. We accept at a 5% level that ¹ = 18:5 . ii As p-value is not < 0:05 we do not reject H0 as there is insufficient evidence to do so. We accept at a 5% level that ¹ = 18:5 . 3

1:2649 p 10

2.947

t¤

r

-2.602

499 ¡ 500

or As the p-value is not (6) As t¤ is not in the critical region we do not reject < 0:01 we do not reject H0 . H0 . (7) We conclude that there is insufficient evidence to suggest that the sample mean is significantly different from the expected value, at a 1% level. (However, we risk making a Type II error by accepting H0 when it is false.)

17:14 ¡ 18:5

) t¤ ¼ ¡1:494

10 9

) t¤ ¼ ¡2:500 is the test statistic (4) We reject H0 if: t¤ lies in the critical or the p-value < 0:01 region (5) or p-value = 2 £ P(T > jt¤ j) 0.005 0.005 = 2 £ P(T > 2:5) ¼ 0:033 86 T

b For º = 15 and ® = 0:01, t® ¼ 2:602 and t ® ¼ 2:947

=

t¤ =

(3)

2

r

p

n sn n¡1

¼ 1:2649 (1) H0 : ¹ = 500 mL and H1 : ¹ 6= 500 mL (2) As ¾ is unknown we use the t-statistic, t(9).

EXERCISE H.3 1 a For º = 15 and ® = 0:05, t® ¼ 1:753 and t ® ¼ 2:131

Now sn¡1 =

185

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IB HL OPT 2ed Calculus

186

WORKED SOLUTIONS (6) As t¤ lies in the or As the p-value < 0:05 critical region we we reject H0 . reject H0 . (7) There is sufficient evidence to reject H0 in favour of H1 . This suggests that at a 5% level we should accept that ¹ 6= 250 and as x was < 250 we surmise that the true ¹ is smaller than 250 mg.

area = 0.02

2.078

g As t¤ ¼ 0:9776 is < 2:078 we have insufficient evidence to reject H0 . So, we reject the claim that ¹ > E95 000. h If the assertion was incorrect, that is, accepting H0 when H1 is correct, we would be making a Type II error. i The 99% confidence interval for the mean income is E92 785 < ¹ < E99 851. This interval confirms that there is not enough evidence to reject H0 as ¹ = 95 000 lies within the interval. Although ® = 0:02, we verify with a 99% confidence interval as we have a one-tailed test.

b The 95% confidence interval for ¹ is 240:7 < ¹ < 244:5 which confirms the above as we are 95% confident that ¹ is well below 250 mg. Hence, we would reject H0 in a and argue again that ¹ < 250 mg. 6 n = 50, x = 26:1, sn = 6:38 As ¾ is unknown, we use sn¡1 as an unbiased estimate of it

50 49

£ 6:38

¼ 6:4448 (1) H0 : ¹ = 24:9 and H1 : ¹ > 24:9 (2) As ¾ is unknown, we use the t-statistic, t(49).

x ¡ ¹0 sn¡1 p n

¼

1 Let X1 represent the test score before coaching, X2 represent the test score after coaching and let U = X2 ¡ X1 . U-values are 5, ¡1, 0, 7, 0, ¡1, 3, 3, 4, ¡1, 1, ¡6

26:1 ¡ 24:9 6:4448 p 50

) t¤ ¼ 1:3166 (4) We reject H0 if: t¤ lies in the critical region (5)

or

the p-value < 0:05

or

p-value = P(T > t¤ ) = P(T > 1:3166) ¼ 0:0970

0.05 T

1.677

U ¼ 1:1667, sn¡1 ¼ 3:4597, n = 12 (1) H0 : ¹ = 0 fno improvementg, H1 : ¹ > 0 (2) As ¾ is not known we use a t-distribution with t(11), and use sn¡1 as an unbiased estimate of ¾. (3) Test statistic is t¤ ¼ 1:168 (4) We reject H0 if: t¤ does not lie in the or the p-value is < 0:05 critical region (5) or p-value = P(T > t¤ ) = P(T > 1:168) 0.05 ¼ 0:1337

1.3166

(6) As t¤ is not in the critical or As the p-value is not region we do not reject < 0:05 we do not H0 . reject H0 . (7) We conclude that, at a 5% level, there is insufficient evidence to suggest that the non-free range chickens have a greater meat protein content.

1.796

or As the p-value is not (6) As t¤ does not lie in the rejection region we do not < 0:05 we do not reject H0 . reject H0 . (7) We conclude that, at a 5% level, there is insufficient evidence to support that there has been improvement.

7 x = E96 318, sn = E14 268, n = 113

=

cyan

£ E14 268

magenta

yellow

0

U ¼ 4:1818 and sn¡1 ¼ 3:2193 (1) H0 : ¹ = 5 and H1 : ¹ 6= 5 (2) As ¾ is unknown we use the t-statistic, t(10). We use sn¡1 as an unbiased estimate of ¾. (3) The test statistic, t¤ =

¼

U ¡ ¹0 sn¡1 p n

4:1818 ¡ 5 3:2193 p 11

¼ ¡0:842 86

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

25

0

5

d e f

113 112

¼ E14 331:55 ) E14 331:55 is an unbiased estimate of ¾. H0 : ¹ = E95 000 and H1 : ¹ > E95 000 As ¾ is unknown and we have to estimate it, the null distribution is a t-distribution with º = n ¡ 1 = 112. t¤ ¼ 0:9776 p-value = P(t > 0:9776) ¼ 0:1652 Critical value is t0:02 ¼ 2:078 as we need to solve P(t > k) = 0:02 or P(t < k) = 0:98 using technology with inverse t.

75

b c

p

2 Let X1 = speed at age 12, X2 = speed at age 13 and let U = X2 ¡ X1 , n = 11 The U-values are: 3, 1, 7, 5, 3, 2, ¡1, 11, 6, 5, 4

n sn n¡1

25

r

a sn¡1 =

T

t¤

95

t¤ =

(3)

EXERCISE H.4

100

p

50

=

n sn n¡1

75

r

and sn¡1 =

T

0

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IB HL OPT 2ed Calculus

WORKED SOLUTIONS (4) We reject H0 if: t¤ does not lie in the critical region (5)

the p-value is < 0:05

or

p-value = 2P(T > jt¤ j) = 2 £ P(T > 0:842 86) ¼ 0:4190

0.025

0.025 -2.228

2.228

ii If ® = 0:01,

or

T

we accept H0 if x < 27 + 2:3263 £ ) x < 28:899 b

= P(X < 28:343 j ¹0 = 29:2)

t¤

= P(X < 28:899 j X » N(29:2, ¼ 0:356

= P(X > 155 j ¹0 = 150) = P(X > 155 j X » N(150,

(2) As ¾ 2 is unknown we use the t-statistic, t(7) and sn¡1 as an unbiased estimate of ¾.

U ¡ ¹0 sn¡1 p n

¼

0:358 p 32 8

(5)

b

or

the p-value < 0:05

or

p-value = P(T > jt¤ j) = P(T > 2:072) ¼ 0:0385 which is < 0:05

0.05 T 1.895

P(Type II error) = P(accepting H0 j H1 is true)

ii When P(Type I error) = P(Type II error), the critical value for x =

150 + 159 2

= 154:5 Check: P(Type I error) = P(X > 154:5 j X » N(150, 4)) ¼ 0:012 22

t¤

P(Type II error) = P(X 6 154:5 j X » N(159, 4)) ¼ 0:012 22 X 3 n = 30, ¹ = unknown, ¾ 2 = 7:5 H0 : ¹ = 37 and H1 : ¹ < 37 a

EXERCISE H.5 a

i

= P(X 6 155 j X » N(159, 4)) ¼ 0:0228

(6) As t¤ lies in the critical or As the p-value is region we reject H0 . < 0:05, we reject H0 . (7) We conclude, at a 5% level, that there is sufficient evidence to support the claim that Type 2 compost improves growth over Type 1 compost.

1

i

a = 0.05 X

0.05 ¾ n

¹0 - z0.05 p

X ¾ n

x > ¹0 ¡ 1:6449 p

If ® = 0:05, we accept H0 if

¾ n p 6 ) x < 27 + 1:6449 £ p 9 x < ¹0 + 1:6449 p

) x > 37 ¡ 1:6449 £

yellow

1 4

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

50

75

25

0

5

95

100

50

75

25

0

100

magenta

p

) x > 36:178 ) accept H0 if x > 36:178, otherwise reject it.

) x < 28:343

cyan

x

¾ n p 7:5 ) x > 37 ¡ 1:6449 £ p 30

¹0 + z0.05 p

x

5

64 )) 16

= P(X > 155 j X » N(150, 4)) ¼ 0:006 21 (about 0:621%)

0:2625 ¡ 0

) t¤ ¼ 2:072 (4) We reject H0 if: t¤ does not lie in the critical region

6 )) 9

2 n = 16, ¹ = unknown, ¾ 2 = 64 H0 : ¹ = 150 and H1 : ¹ > 150 Decision rule: accept H0 if x 6 155 otherwise reject it. a ® = P(Type I error) = P(rejecting H0 j H0 is true)

where U ¼ 0:2625 and sn¡1 ¼ 0:358 32 (1) H0 : ¹ = 0 and H1 : ¹ > 0

t¤ =

6 )) 9

P(Type II error), ® = 0:01 = P(accepting H0 j H1 is true)

ii

3 Let X1 = height using Type 1 compost, X2 = height using Type 2 compost and let U = X2 ¡ X1 . U -values are: 0:2, 0:6, ¡0:2, 0:8, 0:2, ¡0:2, 0:5, 0:2

(3)

2 3

P(Type II error), ® = 0:05 = P(accepting H0 j H1 is true)

i

= P(X < 28:343 j X » N(29:2, ¼ 0:147

(6) As t¤ does not lie in the or As the p-value is not critical region we do not less than 0:05 we do not reject H0 . reject H0 . (7) We conclude, at a 5% level, that there is insufficient evidence to reject the sports commission’s claim.

p

187

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\187IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 3:22:48 PM BRIAN

IB HL OPT 2ed Calculus

188

WORKED SOLUTIONS ) power ¼ 1 ¡ 0:400 ¼ 0:600 c H0 : ¹ = 6 and H1 : ¹ < 6

b True mean is ¹ = 36 ¯ = P(Type II error) = P(accepting H0 j H1 is true) 1 )) 4

= P(X > 36:178 j X » N(36, ¼ 0:361 ) power = 1 ¡ ¯ ¼ 0:639 ( ¼ 63:9%)

0.05 X ¾ n

¹0 - z0.05 p

c If P(Type II error) = 0:1 P(accepting H0 j H1 is true) = 0:1 )

P(X > 36:178 j X » N(¹,

)

P@

0

X ¡¹ p¾

1 )) 4

36:178 ¡ ¹

>

p¾

n

2

)) = P(X > 5:7027 j X » N(6:4, 0:7 15 ¼ 0:999 94 ) power ¼ 0:000 06 (virtually 0).

´

36:178 ¡ ¹ = 0:1 0:5 ³ ´ 36:178 ¡ ¹ P Z6 = 0:9 0:5 36:178 ¡ ¹ ¼ 1:2816 ) 0:5

)

P Z>

)

)

5 Let X = length of a beam (in m). X » N(¹, 0:152 ) a H0 : ¹ = 3:5 m and H1 : ¹ 6= 3:5 m b

36:178 ¡ ¹ ¼ 0:6408 )

¹ ¼ 35:5

0.005

4 ¹ = 6:4, ¾ = 0:7 kg, n = 15, ® = 0:05 a H0 : ¹ = 6 and H1 : ¹ > 6

0.005 X

¾ n

) 3:5 ¡ 2:5758

¹0 + z0.05 p

) x 6 6:2973 otherwise reject it. P(Type II error) = P(accepting H0 j H1 is true) = P(X 6 6:2973 j X » (6:4, ¼ 0:285

x

2

¾ ¾ 6 x 6 ¹0 + z ® p n n 2

¹0 ¡ z ® p

X

Accept H0 if x 6 ¹0 + 1:6449 £

0:7 p 15

³

2

0:15 p 20

´

6 x 6 3:5 + 2:5758

0:72 )) 15

H1 : The die is unfair, so p 6= b X

ii

¾ n

¹0 + z0.025 p

¾ ¾ 6 x 6 ¹0 + z0:025 p n n 6 x 6 6 + 1:960 £

1 ) 4

1 ) 4 1 ) 4

95

100

50

25

0

5

95

yellow

75

iii The test is about the 6:2% level. c If H0 is true, X » B(300, 14 ). For the hypotheses in a we have a two-tailed test with

100

50

1 4

= 1 ¡ P(62 6 X 6 88) with X » B(300, = 1 ¡ [P(X 6 89) ¡ P(X 6 61)] ¼ 1 ¡ 0:971 68 + 0:337 85 ¼ 0:0621

0:72 )) 15

75

25

0

5

95

50

75

25

0

5

95

100

50

75

25

0

5

100

magenta

P(Type I error) = P(Reject H0 j H0 is true) = P(X 6 61 or X > 89 j p = where X » B(300,

0:7 p 15

) 5:6458 6 x 6 6:3542, otherwise reject it. P(Type II error) = P(accepting H0 j H1 is true) = P(5:6458 6 X 6 6:3542 j X » N(6:4, ¼ 0:400

1 4

i A Type I error is rejecting H0 when it is true. This means deciding it is biased when it is in fact fair.

Accept H0 if ¹0 ¡ z0:025 p

cyan

0:152 )) 20

a H0 : The die is fair for rolling a ‘4’, so p =

0.025

0:7 p 15

´

EXERCISE H.6

0.025

) 6 ¡ 1:960 £

0:15 p 20

P(Type II error) = P(accepting H0 j H1 is true) = P(3:4136 6 X 6 3:5864 j X » N(3:4, ¼ 0:343

1

x

³

) 3:4136 6 x 6 3:5864 c

b H0 : ¹ = 6 and H1 : ¹ 6= 6

¾ n

¾ n

¹0 + z ® p

We reject H0 if

0.05

x

¾ n

¹0 ¡ z ® p 2

¹0 - z0.025 p

0:7 p 15

) accept H0 if x > 5:7027, otherwise reject it. P(Type II error) = P(accepting H0 j H1 is true)

A = 0:1

n

³

We accept H0 if x > 6 ¡ 1:6449 £

= 0:1

1

x

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\188IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 3:47:01 PM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS ® ¼ 0:02 and

We reject H0 if X > 4 a If n = 1, X » Po(¸) i P(reject H0 j H0 is true) = P(X > 4 j X » Po(2)) = 1 ¡ P(X 6 3 j X » Po(2)) ¼ 0:143

® ¼ 0:01. 2

Solving P(X 6 k) = 0:01 gives k = 58 P(X 6 k) = 0:99 gives k = 92 The new decision rule is: Roll the die 300 times. If the number of 4s obtained (X) is such that 58 6 X 6 92, accept H0 , otherwise reject it. d If p = 0:32, then X » b(300, 0:32) ) P(Type II error) = P(62 6 X 6 88 j X » B(300, 0:32)) = P(X 6 88) ¡ P(X 6 61) ¼ 0:177

b

i We require that: P(accept H0 j H1 is true) to be < 0:005 ) P(X 6 3 j X » Po(5n)) < 0:005 If n = 1, P(X 6 3 j X » Po(5)) ¼ 0:265 fa iig If n = 2, P(X 6 3 j X » Po(10)) ¼ 0:0103 If n = 3, P(X 6 3 j X » Po(15)) ¼ 0:000 211 which is < 0:005 ) n = 3 minutes is required.

Y » N(¹, 22 ) ) Y » N(¹,

22 ) 16 1 2 (2) )

H0 : ¹ = 330 against H1 : ¹ 6= 330 Critical region is y < 329 or y > 331

329

P(accept H0 j H1 is true) = P(X 6 3 j X » Po(5)) ¼ 0:265

ii

2 Y = volume of drink can (in cm3 ) ) Y » N(¹,

P(rejecting H0 j H0 is true) = P(X > 4 j X » Po(3 £ 2)) = 1 ¡ P(X 6 3 j X » Po(6)) ¼ 0:849

ii

5 Either

Y

331

or

(1)

® = P(Type I error) = P(rejecting H0 j H0 is true)

a

= 1 ¡ P(329 6 Y 6 331 j ¹ = 330) ¼ 0:0455 So ® ¼ 4:55%

50 white

80 black

50 black

H0

H1

We accept H0 if all 4 are black reject H0 if some of the 4 are white. a P(Type I error) = P(reject H0 j H0 is true) = P(not all 4 are black from box (1)) = 1 ¡ P(all 4 are black from (1))

P(Type II error) = P(accepting H0 j H1 is true) = P(329 6 Y 6 331 j Y » N(328, ( 12 )2 )) ¼ 0:022 75

=1¡

a H0 : p = 0:5 (the coin is fair) H1 : p > 0:5 (coin is biased towards heads) b

¼ 0:597

=

d Eri would accept H0 when H1 is true and so risks making a Type II error.

95

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

C350 C150 C4100

+

C450 C050 C4100

¼ 0:249 92 + 0:058 73 ¼ 0:308 65 and 1 ¡ ¯ ¼ 0:691 or 69:1% ) the new decision rule does not give greater power.

4 H0 : ¸ = 2 against H1 : ¸ = 5 We accept H0 if X 6 3

5

C4100

¼ 0:0587

=

yellow

C450 C050

b Power for above test = 1 ¡ ¯ ¼ 1 ¡ 0:0587 ¼ 0:9413 or 94:1% For the new test: P(Type II error) = P(accepting H0 j H1 is true) = P(3 or 4 are black from box (2))

c If p is in fact 0:6 P(Type II error) = P(accepting H0 j p = 0:6) = P(X 6 9 j X » B(12, 0:6)) ¼ 0:917

magenta

C4100

P(Type II error) = P(accepting H0 j H1 is true) = P(all 4 are black from box (2))

i X > 10 is the critical region. ii The significance level (®) of a hypothesis test is the probability of making a Type I error. That is, the probability of rejecting the null hypothesis H0 when it is indeed true. ® = P(rejecting H0 j H0 is true) = P(X > 10 j X » B(12, 12 )) = 1 ¡ P(X 6 9 j X » B(12, 0:5)) ¼ 0:0193

cyan

C480 C020

100

3

(2)

20 white

= P(Y < 329 or Y > 331 j ¹ = 330)

b

189

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\189IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 2:54:01 PM BRIAN

IB HL OPT 2ed Calculus

190

WORKED SOLUTIONS

6 H0 : he rolls a 6, 9 times every 10 rolls ) H1 : p < 0:9 a i Let X = number of 6s in six rolls ) X » B(6, 0:9) P(5 or 6 sixes j p = 0:9) = P(X > 5 j X » B(6, 0:9)) = 1 ¡ P(X 6 4 j X » B(6, 0:9)) ¼ 0:886 = P(X > 5 j X » B(6, ¼ P(X 6 4 j X » B(6,

1 )) 6 1 )) 6

= 1 ¡ P(X 6 3 j X » B(6, ¼ 0:008 70

c

1

a

1 )) 6

8 2

r=

® = P(Type I error) = P(rejecting H0 j H0 is true) = P(S > 38 j m = 3) = 1 ¡ P(S 6 37 j S » Po(3 £ 9)) ¼ 0:0263 ) ® ¼ 2:63%

= = =

a S=

= =

2

H0 : p = 0:25 and H1 : p = 0:38

=

xi is NB(12, p) and the critical region for rejecting

=

magenta

yellow

=

´ ³ y ¡ 10 ´ 10 ³ P xi ¡ 3 i

1 10

1

i=1

P 10

1 20

2

(xi ¡ 3)(yi ¡ 10)

i=1

1 [(¡1 20

£ ¡2) + (¡1 £ 0) + (¡1 £ 2) + (¡1 £ 4)

1 [2 ¡ 2 20 1 [¡10] 20

¡ 4 ¡ 2 ¡ 4]

(xi ¡ x)(yi ¡ y)

i=1 n

P

(xi yi ¡ yxi ¡ xyi + x y)

i=1 n

P

i=1 n

P

i=1 n

P

50

75

25

0

i=1

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

=

P(Type I error) = P(rejecting H0 j H0 is true) = P(S 6 29 or S > 72 j p = 0:25) = P(S 6 29 j p = 0:25) + 1 ¡ P(S 6 71 j p = 0:25)

cyan

x

6

n 1 P (zx )(zyi ) n i=1 i

n P

a

H0 is S 6 29 or S > 72 b

5

= ¡0:5

12

i=1

4

+ (0 £ ¡2) + (0 £ 0) + (0 £ 2) + (1 £ ¡2) + (1 £ 0) + (2 £ ¡2)]

Power = 1 ¡ P(Type II error) = 1 ¡ P(accepting H0 j H1 is true) = 1 ¡ P(S 6 37 j m = 4) = 1 ¡ P(S 6 37 j S » Po(4 £ 9)) ¼ 0:391 (or ¼ 39:1%)

P

3

b There is weak to moderate negative correlation. c x = 3, y = 10 are drawn in a. These lines and the data points within the four quadrants do support the decision in b, as there are 4 data points in the ‘negative’ quadrants and only 1 in the ‘positive’ quadrants. d From technology, ¾ x = 1 and ¾ y = 2

xi

8 X » Geo(p).

y = y = 10

10

95

b

x=x=3

12

The critical region is S 6 37 for H0 acceptance and S > 38 for H1 acceptance. a

y 14

xi yi ¡ y

P

xi ¡ x

P

yi + nx y

xi yi ¡ y(nx) ¡ x(ny) + nx y xi yi ¡ nx y

100

i=1

P(Type II error) = P(accepting H0 j H1 is true) = P(30 6 S 6 71 j p = 0:38) = P(S 6 71 j p = 0:38) ¡ P(S 6 29 j p = 0:38) ¼ 0:999 97 ¡ 0:421 49 ¼ 0:578

EXERCISE I.1

H0 : m = 3 H1 : m = 4

9 P

(0:25)12 (0:75)i¡12

¼ 0:039 03 + 1 ¡ 0:961 54 ¼ 0:0775

P(reject H0 j X » B(6, 0:9)) = P(X 6 3 j X » B(6, 0:9)) ¼ 0:0159 Note: i gives a Type II error ii gives a Type I error

Let S =

11

i=12

ii

7 X » Po(m)

(0:25)12 (0:75)i¡12

71 ¡ ¢ P i¡1

+1¡

¼ 0:000 664

1 )) 6

11

i=12

b H0 : p = 0:9 against H1 : p < 0:9 We will accept H0 if X > 4 and reject H0 if X 6 3. i P(accept H0 j X » B(6, 16 )) = P(X > 4 j X » B(6,

29 ¡ ¢ P i¡1

=

1 )) 6

P(1 six in 6 rolls j X » B(6,

ii

p = 0:9

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\190IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 2:55:25 PM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS n P

n P

(xi ¡ x)2 =

i=1

i=1 n

P

=

i=1 n

P

=

i=1 n

P

= Likewise

n P

i=1

xi2 ¡ 2x

P

n P i=1

n P

=

yi2 ¡ ny 2

4

xi yi ¡ nx y

¶µ

xi2 ¡ nx2

n P

i=1

yi2 ¡ ny 2

84 749 ¡ 12 £ 83:422

1

p

13 126 ¡ 12 £ 32:832

a

5

5

6

7

8

x

a

y

15

p

(100 ¡ 10 £ 32 )(1040 ¡ 10 £ 102 )

10 5

¡10 20

P

xi2 ,

P

1

yi2 ,

y

35

6

b

u v 80

90

x

100

c

u v

P xy = 33 321 P P 2 x = 1001, x = 84 749 P P 2

c From technology,

magenta

95

2 3

3 4

6

7

x

r ¼ 0:982

3 2

5 8

¡1 2

¡3 8

3 ¡4

5 ¡10

ii r ¼ 0:982

7 11 ii r ¼ ¡0:982

¡5 11 ii r ¼ ¡0:982

7 ¡13

100

50

75

25

0

5

95

yellow

5

i For u = 2x + 1, v = ¡3y ¡ 1 u v

100

50

75

25

0

5

95

d

y = 13 126

100

50

75

y = 394,

25

0

5

y = 32:83,

4

i For u = ¡2x + 1, v = 3y ¡ 1

b The data appears to have a strong positive correlation. x = 83:42,

3

i For u = 2x + 1, v = 3y ¡ 1

25

70

1 1

x y

a For

30

60

2

b Negative, as a line of best fit by eye would have negative slope. c From technology, r ¼ ¡0:911 d There appears to be a strong negative correlation between the variables. e As the distance from goal increases, the number of successful shots generally decreases. f Yes, as the distance from the goal does affect the player’s ability to become successful.

40

100

4

20

xi yi , and r.

cyan

3

290 ¡ 10 £ 3 £ 10

= 0:5 Technology gives the values for x, y, n,

P

2

b Using technology, r ¼ 0:816 c There appears to be a moderate positive correlation.

yi2 = 82 + 102 + 122 + 142 + 82 + 102 + 122 + 82

=

95

33 321 ¡ 12(83:42)(32:83)

p

2

¡10 10 £ 40

50

yi2 ¡ ny2

6

= p

75

i=1

8

¶

xi2 = 4 + 4 + 4 + 4 + 9 + 9 + 9 + 16 + 16 + 25 = 4 £ 4 + 3 £ 9 + 2 £ 16 + 25 = 100

) r=

25

n P

10

+ 102 + 82 = 4 £ 82 + 3 £ 102 + 2 £ 122 + 142 = 1040

0

s

¡ nx2

4

P

5

xi2

y

a

xi yi = 16 + 20 + 24 + 28 + 24 + 30 + 36 + 32 + 40 + 40 = 290

P

3

n P

¼ 0:935 confirming a strong positive correlation (Technology could have been used here.)

i=1 n P

s

xi yi ¡ nx y

i=1

i=1

xi2 ¡ nx2

i=1

b

) r=

xi + nx2

xi2 ¡ 2x(nx) + nx2

(yi ¡ y)2 =

sµ

n P

i=1

i=1

Thus r =

12 P

(xi2 ¡ 2xxi + x2 )

191

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\191IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 3:48:39 PM BRIAN

IB HL OPT 2ed Calculus

192

WORKED SOLUTIONS i For u = ¡2x + 1, v = ¡3y ¡ 1

e

¡1 ¡4

¡3 ¡10

ii r ¼ 0:982

5 The product moment correlation coefficient of U and V

¡5 ¡13

=

p

f For u = ax + b and v = cx + d, u and v have the same correlation coefficient, r, as x and y if ac > 0, or have correlation coefficient ¡r if ac < 0.

=

p

u v

=

Cov(U , V )

Var(U )Var(V ) Cov(a + bX, c + dY )

Var(a + bX)Var(c + dY )

E((a + bX)(c + dY )) ¡ E(a + bX) E(c + dY )

p

b2 Var(X)d2 Var(Y )

EXERCISE I.2 1 Cov(X, Y ) = E(XY ) ¡ E(X)E(Y )

=

a Cov(X, X) = E(X 2 ) ¡ [E(X)]2 = Var(X) b

Cov(X, X + Y ) = E(X(X + Y ) ¡ E(X)E(X + Y )) = E(X 2 + XY ) ¡ E(X)E(X + Y )

=

jbdj

jbdj

= E(X 2 ) ¡ [E(X)]2 + E(XY ) ¡ E(X)E(Y ) = Var(X) + Cov(X, Y ) = Cov(X, X) + Cov(X, Y ) ffrom ag

= §1 £

jbdj

p

Var(X)Var(Y )

p

Cov(X, Y )

Var(X)Var(Y )

6 ¾ X2 = 1, ¾ Y2 = 9, ½ =

1 9

From 3, Var(X + Y ) = Var(X) + Var(Y ) + 2Cov(X, Y ) = ¾ X2 + ¾ Y2 + 2½¾ X ¾ Y

= E(X 2 ¡ Y 2 ) ¡ [E(X) + E(Y )][E(X) ¡ E(Y )] = E(X 2 ) ¡ E(Y 2 ) ¡ [E(X)]2 + [E(Y )]2

as ½ =

2

2

½

) Var(X + Y ) = 1 + 9 +

2

= 10 23

=

VarX

=

=

Cov(X, Y )

Var(X)Var(Y )

= =

fE(mX 2 + cX) ¡ E(X)E(mX + c)g2 VarXfm2 Var(X)g

) ½2 = 1

fas m 6= 0g

magenta

yellow

50

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

) ½ = §1

cyan

Var(X + Y )

q

32 3

4 £ 3 1 p 6

q

p

32 3

3 32

b If y = 11, x ¼ 0:511 08 £ 11 + 2:7382 ¼ 8:36 is the estimated score for Mathematics. c If x = 18, y ¼ 0:882 78 £ 18 + 5:5957 ¼ 21:5 is the estimated score for Physics.

95

fmVar(X)g2 m2 [Var(X)]2

p

a From technology, r ¼ 0:671 69 The regression of Y on X is y ¼ 0:882 78x + 5:5957 The regression of X on Y is x ¼ 0:511 08y + 2:7382

1

75

) ½2 =

Var(X + Y )

EXERCISE I.3

fmE(X 2 ) + cE(X) ¡ m[E(X)]2 ¡ cE(X)g2 ) ½ = m2 [Var(X)]2 2

fm(E(X 2 ) ¡ [E(X)]2 )g2 m2 [Var(X)]2

p

1 + 19 (1)(3) 1£

) ½2 =

£1£3

¾ X + ½¾ X ¾ Y ¾X £

fCov(X, mX + c)g2 ) ½2 = Var(X)Var(mX + c) ) ½2 =

Var(X)Var(Y )

100

p

¾

Cov(X, X + Y )

= p

¡ [E(X)]2 ¡ 2E(X)E(Y ) ¡ [E(Y )]2 = E(X 2 ) ¡ [E(X)]2 + E(Y 2 ) ¡ [E(Y )]2 + 2(E(XY ) ¡ E(X)E(Y )) = Var(X) + Var(Y ) + 2Cov(X, Y ) ½=

Cov(X, Y )

Var(X) + Cov(X, Y )

2

= E(X + 2XY + Y ) ¡ [E(X) + E(Y )] = E(X 2 ) + 2E(XY ) + E(Y 2 )

4

p

Var(X)

2

2( 19 )

p

Now the correlation coefficient between X and X + Y

Var(X + Y ) = E((X + Y )2 ) ¡ [E(X + Y )]2

25

Var(X)Var(Y )

Cov(X + Y , X ¡ Y ) = E((X + Y )(X ¡ Y )) ¡ E(X + Y )E(X ¡ Y )

2

0

p

= §½

= E(X ) ¡ [E(X)] ¡ fE(Y ) ¡ [E(Y )] g = Var(X) ¡ Var(Y ) or Cov(X, X) ¡ Cov(Y , Y )

5

Var(X)Var(Y )

bd[E(XY ) ¡ E(X)E(Y )]

=

2

3

p

ac + adE(Y ) + bcE(X) + bdE(XY ) ¡ ac ¡ adE(Y ) ¡ bcE(X) ¡ bdE(X)E(Y )

= E(X 2 ) + E(XY ) ¡ E(X)[E(X) + E(Y )] = E(X 2 ) + E(XY ) ¡ [E(X)]2 ¡ E(X)E(Y )

c If X = c, Cov(XY ) = E(cY ) ¡ E(c)E(Y ) = cE(Y ) ¡ cE(Y ) = 0 for any random variable Y 2

E(ac + adY + bcX + bdXY ) ¡ [a + bE(X)][c + dE(Y )]

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\192IB_HL_OPT-Stat-Prob_an.cdr Thursday, 18 April 2013 5:12:25 PM BRIAN

IB HL OPT 2ed Calculus

193

WORKED SOLUTIONS

t¤ is inside the critical region for ® = 0:05 but outside it for ® = 0:01. (4) At a 5% level, we accept H0 that the data is not correlated. At a 1% level, we do not accept H0 . We accept that the data is correlated. )

d y = 11 is within the given y-values whereas x = 18 is outside the given x-values. As we cannot guarantee that the linear model continues outside the given values, we expect the estimated score for Mathematics to be more accurate. a From technology, r ¼ 0:905 84 The regression line for Y on X is y ¼ 16:960x ¡ 35:528 The regression line for X on Y is x ¼ 0:048 380y +2:7031 b If y = 60, x = 0:048 380 £ 60 + 2:7031 ¼ 5:61 is the estimated cholesterol level.

2

3

2 H0 : ½ = 0 against H1 : ½ 6= 0 n = 10, ) º = df = 8 (1) r ¼ 0:905 84

r

c If x = 5:8, y = 16:960 £ 5:8 ¡ 35:528 ¼ 63 is the estimated resting heart rate.

(2)

a From technology, r ¼ ¡0:940 46 The regression line for Y on X is y ¼ ¡1:2768x + 218:70 The regression line for X on Y is x ¼ ¡0:692 69y+159:72

(3) Either

) t¤ ¼ 6:0481

b When x = 65, y ¼ ¡1:2768 £ 65 + 218:70 ¼ 135:7 sec is the estimated time to swim 200 m breaststroke. 4

a

t¤ ¼ 0:905 84 £

or

i r ¼ 0:784 00 (moderate, positive) ii r ¼ 0:947 74 (strong, positive) iii r ¼ 0:054 72 (very weak, positive)

-t0.025 ¼ -2.306

89

88

T

t0.025 ¼ 2.306

® = 0:01

Men Wt (x)

p-value = 2 £ P(T > 6:0481) ¼ 0:000 307 which is < 0:05 and is < 0:01 ® = 0:05

t¤ ¼ 6.0481

y £ 100% x

b w% =

8 1 ¡ 0:905 842

66

59

93

73

82

77

100

67 t¤ ¼ 6.0481

%BF (w) 13:5 15:9 13:6 16:9 23:7 17:8 15:9 14:3 19:0 17:9 Women Wt (y)

57

68

69

59

62

59

56

66

-t0.005 ¼ -3.355

72

i r ¼ 0:341 81 (weak, positive) ii r ¼ 0:790 75 (moderate, positive) iii r ¼ ¡0:4620 (weak, negative)

3 H0 : ½ = 0 against H1 : ½ 6= 0 (1) r ¼ 0:695 58

r

EXERCISE I.4 1 H0 : ½ = 0 against H1 : ½ 6= 0 n = 10, ) º = df = 8 (1) r ¼ 0:671 69

r

¤

t ¼ 0:671 69 £

(2)

) t¤ ¼ 2:5644 (3) Either

or

t¤ ¼ 0:695 58 £

(2)

) t¤ ¼ 2:5615 (3)

8 1 ¡ 0:671 692

7 1 ¡ 0:695 582

p-value = 2 £ P(T > 2:5615) ¼ 0:0375 which is < 0:05 but > 0:01

(4) At a 5% level we accept H0 that the data is not correlated. At a 1% level we do not accept H0 . We accept that the data is correlated.

p-value = 2 £ P(T < 2:5644) ¼ 0:0334 which is < 0:05 but > 0:01 ® = 0:05

4 H0 : ½ = 0 against H1 : ½ 6= 0 (1) r ¼ 0:606 886

r

t ¼ 0:606 886 £

9 1 ¡ 0:606 8862

) t¤ ¼ 2:291 (3) For ® = 0:05, t ® = t0:025 ¼ 2:262

T

t0.025 ¼ 2.306

¤

(2)

t¤ ¼ 2.5644 -t0.025 ¼ -2.306

T

and for both levels t¤ does not lie in the critical region. (4) At both the 5% level and 1% level we accept H0 , that the data is not correlated.

%BF (w) 29:8 32:4 34:8 30:5 29:0 25:4 28:6 33:3 33:3 c

t0.005 ¼ 3.355

2

® = 0:01 t¤ ¼ 2.5644

t¤

cyan

T

magenta

2.262

T

yellow

95

100

50

75

25

0

t¤ lies in the critical region.

5

95

-2.262

100

50

75

25

0

5

95

t0.005 ¼ 3.355

100

50

75

25

0

5

95

100

50

75

25

0

5

-t0.005 ¼ -3.355

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\193IB_HL_OPT-Stat-Prob_an.cdr Thursday, 18 April 2013 5:48:52 PM BRIAN

IB HL OPT 2ed Calculus

194

WORKED SOLUTIONS For ® = 0:01, t ® = t0:005 ¼ 3:250

8 We test H0 : ½ = 0 against H1 : ½ 6= 0 with n = 20, º = 18

r

2

The t-statistic is t¤ = r t¤

18 1 ¡ r2

For ® = 0:05, t ® = t0:025 ¼ 2:1009 2

3.250

-3.250

r

)

t¤ does not lie in the critical region. (4) So, at a 5% level we do not accept the data is uncorrelated but at a 1% level we do. At a 1% level we accept that ½ = 0 and therefore that X and Y are independent. However, we do not accept this at a 5% level.

r

n¡2 = ¡0:5 1 ¡ r2

r

)

1

a

i

n

= 14¾ 2 ) X1 + 2X2 + 3X3 p deviation ¾ 14. ii

p 0:75 n ¡ 2 > 1:960 p ) n ¡ 2 > 2:613:::: ) n ¡ 2 > 6:829:::: ) n > 8:829:::: ) n>9

= E(X12 ) ¡ 2E(X1 X2 ) + E(X22 ) = ¾ 2 + ¹2 ¡ 2E(X1 ) E(X2 ) + ¾ 2 + ¹2

If n = 11, º = 9, t0:025 ¼ 2:262, t¤ ¼ 2:250 ) t¤ < t0:025

= 2¾ 2 + 2¹2 ¡ 2¹¹ = 2¾ 2

If n = 12, º = 10, t0:025 ¼ 2:228, t¤ ¼ 2:372 ) t¤ > t0:025

2

a Let X be the number of buses needed to get a correct one. X » Geo(0:35) i P(X 6 4) ¼ 0:821

yellow

95

50

75

25

0

ii E(X) =

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

For n = 12 we would reject H0 : ½ = 0 in favour of H1 : ½ 6= 0 ) a sample size of at least 12 is required to conclude, at a 1% level of significance, that X and Y are correlated.

25

has mean 0 and standard

b We use E(X 2 ) = Var(X) + [E(X)]2 E([X1 ¡ X2 ]2 ) = E(X12 ¡ 2X1 X2 + X22 )

If n = 10, º = 8, t0:025 ¼ 2:306, t¤ ¼ 2:121 ) t¤ < t0:025

0

has mean 6¹ and a standard

E(2X1 ¡ 3X2 + X3 ) = 2E(X1 ) ¡ 3E(X2 ) + E(X3 ) = 2¹ ¡ 3¹ + ¹ =0 Var(2X1 ¡ 3X2 + X3 ) = 22 Var(X1 ) + (¡3)2 Var(X2 ) + Var(X3 ) = 4¾ 2 + 9¾ 2 + ¾2 = 14¾ 2 ) 2X1 ¡ 3X2 + X3 p deviation ¾ 14.

If n = 9, º = 7, t0:025 ¼ 2:365, t¤ ¼ 1:984 ) t¤ < t0:025

5

E(X1 + 2X2 + 3X3 ) = E(X1 ) + 2E(X2 ) + 3E(X3 ) = ¹ + 2¹ + 3¹ = 6¹ Var(X1 + 2X2 + 3X3 )

= Var(X1 ) + 22 Var(X2 ) + 32 Var(X3 ) = ¾ 2 + 4¾ 2 + 9¾2

n¡2 1 ¡ 0:62

magenta

4:4139

REVIEW SET A

p = 0:75 n ¡ 2 For infinitely many degrees of freedom the t-distribution approximates the normal distribution Z » N(0, 1) where t ® ¼ z ® = z0:025 ¼ 1:960

cyan

22:4139r2 > 4:4139

least value of jrj is 0:444 (to 3 s.f.)

7 We test H0 : ½ = 0 against H1 : ½ 6= 0 with r = 0:6

2

18r2 > 4:4139 ¡ 4:4139r2

r2 > = 0:196 926 22:4139 p p ) r2 > 0:196 926 ¼ 0:443 76 ) jrj > 0:443 76

6 If r = 1, the t¤ calculation would be t¤ = 1 £ 1¡1 which is undefined. Thus the hypothesis test is unusable. The test is unnecessary as perfect positive correlation exists.

2

r 2 £ 18 > 4:4139 1 ¡ r2

)

which is less than 0:05 but not less than 0:01. So, at a 5% level we reject H0 : ½ = 0 in favour of H1 : ½ 6= 0, and at a 1% level we do not reject H0 . Thus, at a 1% level we accept that ½ = 0 and therefore X and Y are uncorrelated. However, we do not accept this at a 5% level.

If t¤ > 1:960 then

)

)

p-value = 2 £ P(T > 2:16) ¼ 0:0485

The t-statistic is t¤ = 0:6

18 > 2:1009 1 ¡ r2

)

14 ¼ ¡2:16 1 ¡ 0:25

r

jrj

1 1 = ¼ 2:86 (about 3) p 0:35

100

r 5 t¤ = r

jt¤ j > 2:1009

We reject H0 : ½ = 0 if

T

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\194IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 3:07:01 PM BRIAN

IB HL OPT 2ed Calculus

195

WORKED SOLUTIONS b Let Y be the bus that the student will take to school. Y » NB(3, 0:35) i P(Y = 7) =

¡6¢ 2

) G(t) =

(0:35)3 (0:65)4 ¼ 0:115 =

r 3 ii E(Y ) = = ¼ 8:57 p 0:35 )

¡2¢ 2

+

2

¡3¢ 2

a As

Thus, Var(X) = G00 (1) + G0 (1) ¡ [G0 (1)]2 = n(n ¡ 1)p2 + np ¡ n2 p2 = np[(n ¡ 1)p + 1 ¡ np] = np[np ¡ p + 1 ¡ np] = np(1 ¡ p)

P

6 Let X denote the number of hours lost to sickness. X » N(¹, 672 )

c P(X > 1) = P(X = 3 or 5) = 0:2 + 0:2 = 0:4

) X » N(¹,

2

2

2

2

0

= 1¡P@

7

cyan

2

8

375

A

375

where t ® ¼ 1:685 2

a As we have the same group of students, observations are not independent. So, the difference between means is not appropriate. b Student A B C D E F G H I J Pre-test (X1 ) 12 13 11 14 10 16 14 13 13 12 Post-test (X2 ) 11 14 16 13 12 18 15 14 15 11 Difference (D) ¡1 1 5 ¡1 2 2 1 1 2 ¡1 D = X2 ¡ X1 ¹D = 1:1, sn¡1 ¼ 1:8529 fusing technologyg sn¡1 ¼ 1:8529 is an unbiased estimate of ¾. D » t(9) and the 90% confidence interval for D is 0:025 89 < D < 2:1741

yellow

95

50

75

25

0

5

95

100

50

75

25

0

5

95

p67

Using technology, 11:8 < ¹ < 12:8

px (1 ¡ p)n¡x = [p + (1 ¡ p)]n

magenta

¾ ¾ < ¹ < x + t® p n n 2

x ¡ t® p

px (1 ¡ p)n¡x

100

x

50

0

5

95

100

50

75

25

x=0

25

where


787:5) = 1 ¡ P(X 6 787) ¼ 0:514

¡n¢

p67

1 10

b ¾ is unknown, so we use sn¡1 as an unbiased estimate of it. If X is the number of points held by an employee, X » t(39). 95% CI for ¹ is

P(70% prefer right leg) = P(X = 0:7 £ 1050) = P(X = 735) ¼ P(734:5 < X ¤ < 735:5) ¼ 0:000 026 0

75

5


10 and n(1 ¡ p) > 10 ) we can approximate X by a normal variate with

ii

p67

375

ii n = 20 P(X > 15) = 1 ¡ P(X 6 14) ¼ 0:617

i

¡10

= 1 ¡ P(¡2:8903 < Z < 2:8903)

i n = 20 P(X = 14) ¼ 0:169

and

¯

= 1 ¡ P(¡10 < X ¡ ¹ < 10)

4 Let X be the number who prefer right leg kick. X » B(n, 0:75)

¹ = np = 787:5

fCL Theoremg

¯

P(¯X ¡ ¹¯ > 10)

= P(X ¡ ¹ > 10 or X ¡ ¹ < ¡10)

2

= 0:2[3 + 1 + 1 + 3 + 5 ] ¡ 1 = 0:2 £ 45 ¡ 1 =8

a

672 ) 375

We need to find

pi xi2 ¡ (¹X )2 2

for t = 0, 1, 2, 3, ...., n

G00 (t) = n(n ¡ 1)p(1 ¡ p + pt)n¡2 £ p ) G00 (1) = n(n ¡ 1)p2

(0:35) (0:65)2

b ¹X = pi xi = ¡3(0:2) ¡ 1(0:2) + 1(0:2) + 3(0:2) + 5(0:2) = 5(0:2) =1

P

(pt)x (1 ¡ p)n¡x

ii

(0:35)3 (0:65)

pi = 1, 5c = 1 ) c = 0:2

d Var(X) =

x

G0 (t) = n(1 ¡ p + pt)n¡1 £ p = np(1 ¡ p + pt)n¡1 ) G0 (1) = np(1)n¡1 = np ) E(X) = np

¼ 0:235 3

P ¡n¢

px (1 ¡ p)n¡x

i

3

= (0:35)3 [1 + 3 £ 0:65 + 6 £ (0:65)2 ]

P

x

x=0 n

= (1 ¡ p + pt)n b

(0:35)3 +

¡4¢

¡n¢

= (pt + (1 ¡ p))n

the average number of buses until the correct one arrives is approximately 9 buses.

= P(Y = 3 or 4 or 5) =

tx

x=0

P(Y 6 5)

iii

n P

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\195IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 1:39:02 PM BRIAN

IB HL OPT 2ed Calculus

196

WORKED SOLUTIONS c H0 : ¹D = 0 (there is no improvement) H1 : ¹D > 0 (there is an improvement) We perform a 1-tailed test at a 5% level with 9 df and D » t(9). t¤ ¼ 1:8773 and p-value ¼ 0:0466. Since the p-value < 0:05, we reject H0 in favour of H1 . ) at a 5% level of significance, there has been an improvement.

b If ½ = 0, then Cov(X, Y ) = 0 ffrom ag ) E(XY ) ¡ E(X) E(Y ) = 0 fas Cov(X, Y ) = E(XY ) ¡ E(X) E(Y )g ) E(XY ) = E(X) E(Y ) ) X and Y are independent random variables. i Using technology, r ¼ 0:495 94 ii H0 : ½ = 0 against H1 : ½ 6= 0 The test statistic is

c 9 (1) H0 : ¹ = 68 (unchanged) H1 : ¹ < 68 (decreased)

r

¤

t =r

(2) As ¾ is unknown, we use sn¡1 as an unbiased estimate of ¾. r r

n sn = n¡1

sn¡1 =

= 2 £ P(T > 1:713)

=

65 ¡ 68 1:753 p 42

¼ 0:1209 which is not < 0:05 Thus, at a 5% level, we do not reject H0 in favour of H1 . That is, we accept ½ = 0, in which case X and Y are independent random variables.

¼ ¡11:09

We reject H0 if the p-value < 0:05. p-value = 0 ) we reject H0 in favour of H1 . There is sufficient evidence at a 5% level to accept H1 that the diet and exercise program has reduced Yarni’s resting pulse rate.

12

a

Var(X)

µ

1 P = Var Xi n i=1

7:55 ¡ 7:82 1:83 p

z¤

µ ½µ

¼ ¡1:022

1 = n 1 = n

0.025

1 = n

Z

2

magenta

yellow

Xi

= E@ i=1

A

n

¶

µ

n P 1 = E Xi n i=1

=

n 1 P E(Xi ) n i=1

=

1 £ n¹ n

½ µ E

½

i=1 n P

i=1 n P

¶ Xi

Xi

2

¡ nX

¶ 2

n P

2

¾¶

2

¾

¡ nE(X )

2

i=1

½

n P

2

¾

E(Xi ) ¡ nE(X )

¾ 2

i=1

2

(Var(Xi ) + [E(Xi )] ) ¡ n(Var(X) + [E(X)] )

1 n

½

n P

µ

(¾ 2 + ¹2 ) ¡ n

i=1

¾2 + ¹2 n

¶¾

50

ª 1© = n(¾ 2 + ¹2 ) ¡ ¾ 2 ¡ n¹2 n ¾2 = ¾ 2 + ¹2 ¡ ¡ ¹2 n ³ ´ 1 = ¾2 1 ¡ n

25

0

=

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

The p-value is not less than 0:05. ) we do not reject H0 . (7) There is insufficent evidence to reject H0 at a 5% level of significance. So, we accept that there has been no change in the colony’s mean weight at this level.

25

or

75

(6) z ¤ does not lie in the critical region. ) we do not reject H0 .

0

1

fsince Var(U) = E(U 2 ) ¡ [E(U)]2 for any random variable Ug

-1.960 = z ®

5

1 n

=E p-value is < 0:05 p-value = 2 £ P(Z 6 ¡1:022) ¼ 0:3067

or or

0

n P

=¹ E(Sn2 )

c

48

cyan

1 P Var(Xi ) n2 i=1

p¾

(4) We reject H0 if: z ¤ is in the critical region (5)

0.025

0

x ¡ ¹0 n

) z¤ =

E(X)

=

(2) As ¾ is known (¾ = 1:83) we use the Z-distribution (the null distribution). (3) The test statistic is z =

b

¶

1 £ n¾ 2 n2 ¾2 = n

10 (1) H0 : ¹ = 7:82 (no change) H1 : ¹ = 6 7:82

¤

n

n

=

95

sn¡1 p n

9 1 ¡ 0:495 942

p-value = 2 £ P(T > jt¤ j)

100

(4) (5) (6) (7)

x ¡ ¹0

r

n¡2 = 0:495 94 1 ¡ r2

) t¤ ¼ 1:713 with º = n ¡ 2 = 9

42 £ 1:732 41

) sn¡1 ¼ 1:753 The null distribution is t(41). (3) The test statistic is t¤ where t¤ =

Cov(X, Y ) ¾X ¾Y

a By definition, ½ =

11

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\196IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 10:20:43 AM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS =

³n ¡ 1´ n

2 X » B(1000,

¾2

a

6 ¾2 = ) Sn2 is a biased estimator of ¾ 2 .

³ n ´

2 d E(Sn¡1 )=

2 Sn¡1

)

1

a

i

b

= ¾2 is an unbiased estimator of ¾ 2 .

REVIEW SET B

i E(X) = np = 1000 £ = 600

P

Since pi = 1, 0:3 + 0:2 + 0:2 + P(X = 6) = 1 ) P(X = 6) = 0:3

¼ P(a ¡

0 ¡3 ¡8 ¡5 (0:3)

4 1 ¡4 ¡1 (0:2)

P(X = x) = ) P(X = k + r ¡ 1) =

P(X = k) =

8 5 0 3 (0:2)

11 8 3 6 (0:3)

game 1

4

1 0:06

) G(t) =

X +Y prob.

3 0:15

4 0:04

5 0:06

8 0:13

11 0:06

=

0

p(1 ¡ p)k¡1

yellow

1 P

tx p(1 ¡ p)x¡1

1 p P [t(1 ¡ p)]x 1 ¡ p x=1

p u1 £ provided jrj < 1 1¡p 1¡r p t(1 ¡ p) = £ provided jt(1 ¡ p)j < 1 1 ¡ p 1 ¡ t(1 ¡ p) pt 1 provided jtj < = 1 ¡ t(1 ¡ p) 1¡p p [1 ¡ t(1 ¡ p)] ¡ pt(¡1 + p)

=

G0 (t) =

=

[1 ¡ t(1 ¡ p)]2 p (1 ¡ t + pt)2

b E(X) = G0 (1) =

p

95

100

50

75

[1 ¡ 1 + p]2 p = 2 p 1 = p

25

0

for k = 1, 2, 3, 4, ....

x=1

5

95

100

50

75

25

0

5

95

100

50

75

25

¡ k¡1 ¢

pr (1 ¡ p)k¡1

a If X » Geo(p), then X has PDF P(X = x) = p(1 ¡ p)x¡1 for x = 1, 2, 3, 4, ....

0 0:16

0

r¡1

P(X = k) = p(1 ¡ p)k¡1 ) NB(1, p) = Geo(p)

¡3 0:09

5

95

¡ k+r¡2 ¢

= (1 ¡ p)k¡1 p for k = 1, 2, 3, 4, 5, .... But for X » Geo(p),

¡4 0:10

100

50

75

25

0

p (1 ¡ p)k+r¡1¡r

r¡1

Letting r = 1, gives

For example: P(X + Y = ¡8) = P(X = ¡5, Y = ¡3) = 0:3 £ 0:5 = 0:15 P(X + Y = ¡4) = P(X = ¡1, Y = ¡3) = 0:2 £ 0:5 = 0:1, and so on E(X + Y ) = ¡8(0:15) ¡ 4(0:10) ¡ 3(0:09) + :::: + 11(0:06) = 0:8 or E(X + Y ) = E(X) + E(Y ) = 0:7 + 0:1 = 0:8 ) expected return = 80 cents/game iii Organiser’s gain = 1 £ (500 + 500 + 1000) ¡ 0:7 £ 500 ¡ 0:1 £ 500 ¡ 0:8 £ 1000 = $800

5

¡ x¡1 ¢ r p (1 ¡ p)x¡r r¡1 ¡ k+r¡1¡1 ¢ r

for k = 1, 2, 3, 4, 5, .... =

¡8 0:15

magenta

6 X2 6 b + 12 ) where X2 is normal

3 If X » NB(r, p) then

X +Y prob.

cyan

1 2

f“the continuity correction”g

i E(Y ) = ¡3(0:5) + 2(0:3) + 5(0:2) = 0:1 ) expected return is 10 cents/game. ii game 2 X + Y values 5 2 ¡3

= 1000 £ 35 £ 25 = 240 p = 240 ¼ 15:5

P(580 6 X 6 615) = P(X 6 615) ¡ P(X 6 579) = 0:841 47 ¡ 0:093 12 ftechnologyg ¼ 0:7484

P

(0:2) (0:3) (0:5)

ii Var(X) = np(1 ¡ p) 3 5

c P(579:5 6 X 6 615:5) ¼ 0:7486 ftechnologyg d If X » B(n, p) where np > 10 and n(1 ¡ p) > 10 then X » N(np, np(1 ¡ p)) and P(a 6 X1 6 b) where X1 is binomial

ii E(X) = pi xi = ¡5(0:3) ¡ 1(0:2) + 3(0:2) + 6(0:3) = 0:7 ) expected return is 70 cents/game. iii At 50 cents/game, they would on average lose 20 cents each game. However, if they charge $1/game, they would expect to gain 30 cents each game. b

3 ) 5

) ¾X

E(Sn2 )

n¡1 n n¡1 2 £ ¾ = n¡1 n

197

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\197IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 10:33:52 AM BRIAN

IB HL OPT 2ed Calculus

198

WORKED SOLUTIONS G0 (t) = p(1 ¡ t + pt)¡2 00

f P(X > 1:3) = 1 ¡ P(X 6 1:3) ¡3

G (t) = ¡2p(1 ¡ t + pt)

¡

= 1 ¡ 1 ¡ e¡0:8£1:3

£ (¡1 + p)

= e¡1:04 ¼ 0:353

= ¡2p(1 ¡ t + pt)¡3 (p ¡ 1) ) G00 (1) = ¡2p(p)¡3 (p ¡ 1)

2(1 ¡ p) p2

=

£

6 Let X be the volume of a bottle in mL. X » N(376, 1:842 )

¤2

Now Var(X) = G00 (1) + G0 (1) ¡ G0 (1)

Then X is the average volume of each sample of 12.

³

2(1 ¡ p) 1 1 = + ¡ 2 p2 p p 2 ¡ 2p + p ¡ 1 = p2 1¡p = p2 5

X » N 376,

1:842 12

´

a P(X < 373) ¼ 0:0515 ) about 5:15% will have a volume less than 373 mL. b P(X < 375) ¼ 0:0299 ) about 3% of all packs of 12 will have an average contents less than 375 mL. c From a and b there is a smaller chance of picking a 12-pack that does not meet the rules than that for an individual bottle. Hence, would prefer method II.

a, d f (x) = 0:8e¡0:8x , x > 0

d Let X » N(¹, 1:842 ) We want P(X < 375) = 0:01

y 0.8

0

) P@

1

X ¡¹ 1:84 p


502 j X » N 500,

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\199IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 1:41:33 PM BRIAN

Var(X2 )

IB HL OPT 2ed Calculus

200

WORKED SOLUTIONS 9 4 Var(X1 ) + 25 25 9 2 4 2 ¾ + 25 ¾ 25 13 2 ¾ 25 52 2 ¾ which is 100

Var(T2 ) = = = =

Var(X2 )

c

i Let Y denote returns based on numbers of errors. y P(Y = y)

< Var(T1 )

) Eve is not correct. T2 is a more efficient estimator than T1 .

ii E(Y ) =

P

$10 q

¡$8 1 ¡ q + q ln q

$1 ¡q ln q

yi pi

= $ [10q ¡ q ln q ¡ 8(1 ¡ q + q ln q)]

a b E(X1 ) + E(X2 ) c c a b = ¹+ ¹ c c ³a + b´ = ¹ c

= $[18q ¡ 9q ln q ¡ 8]

d E(T3 ) =

iii To receive a bonus (positive) we require 18q ¡ 9q ln q ¡ 8 > 0. First we solve 18q ¡ 9q ln q ¡ 8 = 0 ) q ¼ 0:268 For q = 0:267, f (q) ¼ ¡0:0208 < 0 For q = 0:268, f (q) ¼ 0:000 045 1 which is > 0 ) required q = 0:268

=¹ ) a+b=c

REVIEW SET C 1

Z

a Y = 2X3 ¡ 2X2 ¡ X1 i E(Y ) = 2E(X3 ) ¡ 2E(X2 ) ¡ E(X1 ) = 2a ¡ 2(3) ¡ (2) = 2a ¡ 8

4

0

for a well-defined pdf, and that

Z

Var(Y )

ii

¡1¢

16 1 1 + 4 8

= 4b +

1 8

+

If Var(Y ) = 1, 4b +

3 8

0

5 8 5 32

Z From (2), 0

)

¼ 0:106 1 6

and ¹X =

1 1 = =6 1 p

)

a X » Po(m) where m is the unknown mean number of errors per page.

a As ¹ =

5

7 10

a b + ¡0= 5 4

7 10

0

1

(¡4x3 + 6x2 ) dx

pr , ¹ ¡ ¹p = pr 1¡p

) p=

yellow

95

¹ ¹+r

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

=

) ¹ = p(¹ + r)

= 1 ¡ q + q ln q

50

¸1

ax5 bx4 + 5 4

) ¹ = pr + ¹p

= 1 ¡ [P(X = 0) + P(X = 1)]

75

·

¼ 0:0998 the service station runs out of petrol about 10% of the time.

ii P(X > 1) = 1 ¡ P(X 6 1)

25

(ax4 + bx3 ) dx = 0:7

0:95

mx e¡m for x = 0, 1, 2, 3, .... x! me¡m i P(X = 1) = = q(¡ ln q) = ¡q ln q 1!

0

) 3a + 4b = 12 .... (3) 1

Z

b P(X = x) =

5

a b + ¡0=1 4 3

=

So, on average a player takes 6 rolls to win E10. As Pierre wants to make E2/game, he must charge E12 over the 6 rolls, which is E2 per roll.

magenta

=1 0

) 4a + 5b = 14 .... (4) Solving (3) and (4) simultaneously gives a = ¡4, b = 6. b P(runs out of petrol) = P(X > 0:95)

6

cyan

¸1

ax4 bx3 + 4 3

)

Let X be the number of rolls needed to obtain a ‘six’.

3

·

)

d P(X3 > 8b) = P(X3 > 1:25)

6

(ax3 + bx2 ) dx = 1

)

c Y is a linear combination of normal variables. Consequently Y is Normal. So, Y = N(0, 1).

X » Geo

1

=1

) b=

)

Z

From (1),

) 4b =

¡1¢

x f (x) dx = 0:7 .... (2) 0

b If E(Y ) = 0, 2a ¡ 8 = 0 ) a=4

2 P(a ‘six’) =

1

E(X) =

= 22 Var(X3 ) + (¡2)2 Var(X2 ) + (¡1)2 Var(X1 ) = 4b + 4

1

f (x) dx = 1 .... (1)

a We know that

black

Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\200IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 10:52:00 AM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS

0

µ

¹t 1 ¡ ¹+r

¹+r¡¹ ¹ + r ¡ ¹t

=

µ =

à =

r ¹ + r ¡ ¹t

µ c As r ! 1,

¶r

1 ¹(1¡t) r

1+

¹(1 ¡ t) r

1+ f lim

³

q

!r

n t(0:025, n ¡ 1) n¡ s 1 n = 3:79 ) p n t(0:025, n ¡ 1) £ 11:2 ) p = 3:79 n¡1

´r

p ) t(0:025, n ¡ 1) = 0:3384 n ¡ 1 which is not solvable algebraically as n must be known for a t calculation.

¶r ! e¹(1¡t)

n

·

d n¡1=

! ea , ¹ and t fixedg

·

lim G(t) = e¡¹(1¡t)

)

r!1

) n=

¹(t¡1)

=e which is the PGF for the Poisson variable with mean ¹.

³

m d Po(m) = lim NB r, r!1 m+r

a X » N(63:4, 40:1) P(X 6 60) ¼ 0:296 About 29:6% of sausages produced weigh 6 60 grams.

¡

b W » N 63:4,

40:1 10

¢

P@

W ¡ 63:4

q

40:1 n

1

6

60 ¡ 63:4

q

0

) P @Z 6

)

40:1 n

40:1 n

7

The test statistic is t¤ =

A = 0:01

Note: We do not have enough information to determine the probability of making this type of error. b A 90% confidence interval for the mean difference is ]¡0:121, 10:121[ and as the null hypothesis value of ¹D = 0 is within the CI, then at a 5% level, this is consistent with the acceptance of H0 .

a As ¾ is unknown, the t-distribution replaces the Z-distribution for the CI and sn¡1 is used as an unbiased estimate of ¾. Thus, the 95% CI for ¹ is:

yellow

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

5¡0

8:2614 p 9

² as p-value > 0:05 or ² as t¤ does not lie in the rejection region (t > 1:860) then we do not reject H0 and are subject to making a Type II error, that is, accepting H0 when it is in fact false.

sn¡1 sn¡1 6 ¹ 6 x + t(0:025, n ¡ 1) p n n

0

=

which is > 0:05

x ¡ t(0:025, n ¡ 1) p

5

p¾ n

The decision:

p n ¼ 4:333

magenta

d ¡ ¹0

and p-value ¼ 0:0535

) n ¼ 18:77 samples of size 19 are needed.

cyan

+1

) t¤ ¼ 1:816

p ¡3:4 n p ¼ ¡2:3263 40:1 )

)

A = 0:01 1

¡3:4

q

¸2

We consider D = X2 ¡ X1 a H0 : ¹D = 0 against H1 : ¹D > 0 (the course was effective) D values are: 12, 9, 18, ¡3, ¡9, 4, 0, 10, 4 where d = 5 and sn¡1 ¼ 8:2614 As ¾ is unknown, we use sn¡1 as an unbiased estimate of ¾.

P(W 6 60) = 0:01

We require

0

t(0:025, n ¡ 1) 0:3384

¸2

8 Let X1 be the number of fish caught before the course, and X2 be the number of fish caught after the course.

fCL Theoremg

c P(W 6 60) ¼ 0:0448 In approximately 4:48% of all samples of 10 sausages, the mean weight in the sample will be 6 60 g. d

t(0:025, n ¡ 1) 0:3384

If n = 33, n ¼ 37:23 If n = 34, n ¼ 37:14 If n = 35, n ¼ 35:06 X If n = 36, n ¼ 36:90 Thus n = 35.

´

95

6

sn¡1 = 147:49 n t(0:025, n ¡ 1)sn¡1 = 3:79 p n

x + t(0:025; n ¡ 1) p )

´ a n

1+

n!1

c In order to find n we would need to solve

¶r

1 1+ ¹ (1 ¡ t) r

³

=

b Adding the two outer boundary limits gives 2x. ) 2x = 139:91 + 147:49 = 287:4 ) x = 143:7

A

100

b Hence,

G(t) = @

1r

¹ 1 ¡ ¹+r

201

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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\201IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 10:55:05 AM BRIAN

IB HL OPT 2ed Calculus

202

WORKED SOLUTIONS

9 (1) H0 : ¹ = 200 and H1 : ¹ < 200 (2) The null distribution is Z. (3) The test statistic is z ¤ where z¤ =

x ¡ ¹0 p¾

=

1 ¡ r2 > 0,

Since

) 25r2 < 4:2436 ¡ 4:2436r2 ) 29:2436r2 < 4:2436

196:4 ¡ 200

) r2 < 0:145 11 p ) r2 < 0:380 93 ::::

11:36 p 35

n

) z ¤ ¼ ¡1:875 (4) We reject H0 if z ¤ lies in the critical region. (5)

) jrj < 0:380 93 :::: Thus for X, Y independence, the greatest value of jrj is 0:3809 (to 4 s.f.).

z¤

2 a Using Sn¡1 =

12

n Sn2 n¡1

Sample A gives Z

0

2 Sn¡1 = 2 Sn¡1 =

b

= = )

cyan

1 Let S be the volume of a small bottle and let L be the volume of a large bottle. ) S » N(338, 32 ) and L » N(1010, 122 ) a Consider U = L ¡ (S1 + S2 + S3 ) = L ¡ S1 ¡ S2 ¡ S3 ) E(U) = E(L) ¡ E(S1 ) ¡ E(S2 ) ¡ E(S3 ) = 1010 ¡ 3 £ 338 = ¡4 mL

50

25

0

5

95

50

100

yellow

95

and Var(U) = Var(L) + Var(S1 ) + Var(S2 ) + Var(S3 ) = 122 + 3 £ 32 = 171 mL2 P(L > S1 + S2 + S3 ) = P(L ¡ S1 ¡ S2 ¡ S3 > 0) = P(U > 0) where U » N(¡4, 171) ¼ 0:380

75

25

0

5

95

100

50

6= ¾ T1 is a biased estimator of ¾ 2 , and so t1 is a biased estimate of ¾ 2 .

REVIEW SET D

25r2 < 4:2436 1 ¡ r2

magenta

8 30 8 30 28 ¾ 30 2

) 77a + 84b = 88c

25 < 2:06 1 ¡ r2

75

25

0

5

95

100

50

75

25

0

5

)

SA2 +

¢

22 2 S 30 B 2 22 E(SA ) + 30 E(SB2 ) £ 78 ¾ 2 + 22 £ 21 ¾2 30 22 2 30

100

r

we accept H0 when r

¡8

a b E(SA2 ) + E(SB2 ) c c a b 7 2 ¾2 = £ 8 ¾ + £ 21 c c 22 ³ 7a 21b ´ = + ¾2 8c 22c 7a 21b which is ¾ 2 if + =1 8c 22c

) 2 £ P(T > jt¤ j) > 0:05 ) P(T > jt¤ j) > 0:025 ) P(T 6 jt¤ j) < 0:975 ) t¤ ¼ 2:060 ftechnologyg

)

£ 5:1 ¼ 5:34

c E(T2 ) =

n = 27 and so º = 27 ¡ 2 = 25 We require the p-value to be > 0:05 for H0 acceptance.

n¡2 1 ¡ r2

22 21

8(4:2) + 22(5:1) = 4:86 30 =

11 H0 : ½ = 0 (X and Y are independent) against H1 : ½ = 6 0

r

i t1 =

ii E(T1 ) = E

c A Type II error is accepting H0 when H0 is in fact false. This means accepting m = 3 when in fact m = 3:4. P(Type II error) = P(34 6 S 6 56 j S » Po(15 £ 3:4)) = P(S 6 56 j S » Po(51)) ¡ P(S 6 33 j S » Po(51)) ¼ 0:782 27 ¡ 0:004 79 ¼ 0:777

But t¤ = r

£ 4:2 = 4:8

So, 4:8 and 5:34 are unbiased estimates of ¾ 2 .

a X is a discrete random variable with values 0, 1, 2, 3, 4, .... ) S is also a discrete random variable and the critical region is f0 6 S 6 33g [ fS > 57g. b A Type I error is rejecting H0 when H0 is in fact true. This means, deciding that m 6= 3 when in fact m = 3 (and X » Po(3)). P(Type I error) = P(S is in the critical region j S » Po(15 £ 3)) = P(0 6 S 6 33 or S > 57 j S » Po(45)) = P(S 6 33 j S » Po(45)) + 1 ¡ P(S 6 56 j S » Po(45)) ¼ 0:038 339 + 1 ¡ 0:952 745 ¼ 0:0856

10

8 7

Sample B gives

(6) Since z ¤ does not lie in the critical region we do not reject H0 . (7) There is insufficient evidence to reject H0 , that is, there is insufficient evidence to suggest that her present mean has fallen to less than 200.

75

-z0.02 = -2.054

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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\202IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 11:04:03 AM BRIAN

IB HL OPT 2ed Calculus

203

WORKED SOLUTIONS b Consider V = L ¡ 3S ) E(V ) = E(L) ¡ 3 E(S) = 1010 ¡ 3 £ 338 = ¡4 and Var(V ) = Var(L) + 9 Var(S) = 122 + 9 £ 32 = 225 ) V » N(¡4, 225) P(L > 3S) = P(L ¡ 3S > 0) = P(V > 0) ¼ 0:395 2

When t = 1,

) ¹ = G0 (1) = (® + 1)¯

3 4

)

·

P(X = 5) =

= (® + 1)¯

E(Y ) =

G00 (1)

)

= (® + 1)¯ 2 n P

of 14 = 7

6

a

Thus n( 14 ) » B(32, 14 )

i X

sn2

=

r

c

=

n

230 ¼ 15:33 15

n sn ¼ 4:0532 is an unbiased estimate n¡1

of ¾. i A CI for ¹ is 124:94 < ¹ < 129:05 and x for this sample is the midpoint of the CI. )

3 4 1 4

=8

(xi ¡ x)2

i=1

b sn¡1 =

8(1 ¡ p) = 6 ) p=

¤2

= (® + 1)¯ 2 [® + 2 ¡ ® ¡ 1] = (® + 1)¯ 2 £ 1

10:55 e¡10:5 ¼ 0:0293 5!

) 1¡p =

¸

= (® + 1)¯ 2 (® + 2) ¡ (® + 1)¯ + (® + 1)¯ ¡ (® + 1)2 ¯ 2

Var(X) = np(1 ¡ p) = 6,

b

¯(® + 2) ¡ 1 = (® + 1)¯ 1

£

As E(X) = np = 8 and

a

·

¸

and Var(X) = G00 (1) + G0 (1) ¡ G0 (1)

Y » Po(7) ) P(Y < 7) = P(Y 6 6) ¼ 0:450 3

= (® + 1)¯

¡¯

fquotient ruleg

b Let Y be the number of patients arriving between 10:00 am and 10:30 am. 1 2

1

(¡®¡2)(1¡¯ ln t)¡®¡3 £ t ¡(1¡¯ ln t)¡®¡2 £1 t2

of 14 = 10:5

X » Po(10:5)

h1i

G00 (t)

b

a Let X be the number of patients arriving between 9:00 am and 9:45 am. E(X) =

ln t = 0

x=

124:94 + 129:05 = 126:995 2

As ¾ is unknown (had to be estimated), we have a t-distribution with º = 15 ¡ 1 = 14, that is, t » t(14). A 95% CI is 124:75 < ¹ < 129:24

) n = 32

) P(X > 4) = 1 ¡ P(X 6 3) ¼ 0:975

ii t¤ =

ii As n > 30, we can use the Central Limit Theorem. 6 X » N(8, 32 )

jx ¡ ¹j p¾ n

¼

j126:995 ¡ 129:05j 4:0532 p 15

) t¤ ¼ 1:9636 :::: and P(T < 1:9636 ::::) ¼ 0:965 12

P(7:9 6 X 6 8:1) ¼ 0:406

)

4 As f (x) is the normal PDF,

Z

1

1 1 ¡ e 2 p 2¼¾

¡1

Z

1

)

1 ¡2

¡ x¡¹ ¢2 dx = 1

¾

¡ x¡¹ ¢2 dx =

¾

e

p

0.03488

2¼¾

f¾ is fixedg

0

¡1

1

¡2

¡ x¡9 ¢2 2

e ¡1 1

Z )

1

¡ 8 (x¡9)2

e

) ® ¼ 2 £ 0:3488 ) ® ¼ 6:98 ) we have a 7% confidence level.

p dx = 2 2¼ p dx = 2 2¼

7

a B, as: ² A is closer to normality ² A’s standard deviation is much less than 9:21

¡1

5 G(t) = (1 ¡ ¯ ln t)

, ® and ¯ are parameters

0

a G (t) = (¡® ¡ 1)(1 ¡ ¯ ln t)¡®¡2 £ ¡¯

·

cyan

¾2 = X

magenta

yellow

50

75

25

0

5

95

100

50

75

25

0

¾2 9:212 = ¼ 9:425 n 9

c P(X > 12) ¼ 0:423

5

95

100

50

¸

b ¹X = ¹ = 11:4

t

t

75

25

0

5

95

100

50

75

25

0

5

= (® + 1)¯

(1 ¡ ¯ ln t)¡®¡2

³1´

95

¡®¡1

100

1

T

0.96512

Letting ¹ = 9 and ¾ = 2 we get

Z

1.9636

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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\203IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 11:08:30 AM BRIAN

IB HL OPT 2ed Calculus

204

WORKED SOLUTIONS d Firstly we need to consider P(¹Y ¡ 3 6 Y 6 ¹Y + 3) = 0:95

µ

) P

a As X » Geo(0:25), S » NB(12, 0:25) b H0 should be accepted if 32 6 S 6 74 fS > 12g.

9

¶

¹Y ¡ 3 ¡ ¹Y

c ® = P(Type I error)

Y ¡ ¹Y ¹ + 3 ¡ ¹Y 6 6 Y = 0:95 ¾Y ¾Y µ ¶ ¡3 3 ) P 6Z6 = 0:95 ¾Y ¾Y

¾Y

= P(rejecting H0 j H0 is in fact true) = P(S 6 31 or S > 75 j S » NB(12, 0:25)) = P(S 6 31 j S » NB(12, 0:25)) + 1 ¡ P(S 6 74 j S » NB(12, 0:25)) 31 ¡ ¢ P i¡1

=

i=12

0.95

0.025

0.025

+1 ¡

11

(0:25)12 (0:75)i¡12

74 ¡ ¢ P i¡1 11

i=12

) P

Z6¡

3 ¾Y

) ¡

Z

3 ¾Y

¶

¼ 0:064 426 + 1 ¡ 0:974 523 ¼ 0:089 90 Thus ® ¼ 0:9 d P(Type II error)

= 0:025

= P(accepting H0 j H1 is true)

3 ¼ ¡1:960 ¾Y

= P(32 6 S 6 74 j S » NB(12, 0:2)) = P(S 6 74 j S » NB(12, 0:2))

) ¾ Y ¼ 1:531

¡ P(S 6 31 j S » NB(12, 0:2))

¾Y n

But ¾ Y = p ) n=

=

¾ Y2

i=12 31

¾2

i=12

) n ¼ 36:2 So, n must be at least 37.

fsn¡1 = =

p

a A Type I error would result if it was determined that Quickchick is supplying underweight chickens when they are in fact not. b A Type II error would result if Quickchick is supplying underweight chickens when it is determined that they are not.

11

a

563 ¡ 546 59:049 p 50

n sn n¡1 50 49

£

6

) t¤ ¼ 2:0357 (4) We reject H0 if: t¤ is in the critical region

or

the p-value is < 0:05

(5)

or

p-value = P(T > 2:0357) ¼ 0:0236

4 2

10

T

0 t0.05 ¼ 1.677

yellow

50

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

5

95

100

50

75

25

0

or As the p-value is (6) As t¤ lies in the critical region, we reject H0 . < 0:05, we reject H0 . (7) There is sufficient evidence to reject H0 in favour of H1 . We accept that the new brand has longer life at a 5% significance level.

magenta

20

30

40

50

60

70

80

x

b r ¼ ¡0:9462 ftechnologyg c The correlation is strong and negative. d Yes, as the distance from the target does affect the number of hits. e y = ¡0:1134x + 11:27 hits f When x = 50, y = ¡0:1134 £ 50 + 11:27 = 5:6 So, on average we expect about 5:6 hits from a 50 m distance, for every 10 arrows shot. ) for 100 arrows we predict 5:6 £ 10 = 56 hits.

0.05

75

t¤

5

y 10 8

p 3417

¼ 59:049g

cyan

(0:2)12 (0:8)i¡12

10

95

r

=

11

¼ 0:818 ) power ¼ 1 ¡ 0:818 ¼ 0:182 or 18:2%

(2) As ¾ 2 is unknown we use the t-distribution with º = 49. (3) The test statistic is t¤ where sn¡1 p n

(0:2)12 (0:8)i¡12

¼ 0:830 84 ¡ 0:012 72

8 (1) H0 : ¹ = 546 (there is no change) H1 : ¹ > 546 (there is an increase)

x ¡ ¹0

11

P ¡ i¡1 ¢

¡

Y

t¤ =

74 ¡ ¢ P i¡1

100

µ

¡3 ¾Y

(0:25)12 (0:75)i¡12

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Y:\HAESE\IB_HL_OPT-Stat-Prob\IB_HL_OPT-Stat-Prob_an\204IB_HL_OPT-Stat-Prob_an.cdr Friday, 19 April 2013 1:43:11 PM BRIAN

IB HL OPT 2ed Calculus

WORKED SOLUTIONS

205

g No, as 100 m is outside the data set and so any calculation would be unreliable. h x ¼ 7:894y + 93:68 metres 12

a E(T1 ) = =

2 E(X1 ) + 13 3 2 ¹ + 13 ¹ 3

E(X2 )

=¹ E(T2 ) = a E(X1 ) + (1 ¡ a) E(X2 ) = a¹ + (1 ¡ a)¹ = a¹ + ¹ ¡ a¹ =¹ ) T1 and T2 are both unbiased estimators of ¹. b Var(T1 ) = = = Var(T2 ) =

4 Var(X1 ) + 19 Var(X2 ) 9 4 2 ¾ + 19 ¾ 2 9 5 2 ¾ 9 2 a Var(X1 ) + (1 ¡ a)2 Var(X2 )

= a2 ¾ 2 + (1 ¡ a)2 ¾ 2 = (2a2 ¡ 2a + 1)¾ 2 c If T2 is a more efficient estimator than T1 Var(T2 ) < Var(T1 ) ) 2a2 ¡ 2a + 1 < ) a2 ¡ a + ) (a ¡

2 )(a 3

2 9 1 ) 3

¡